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AN 


ALGEBRA 


FOR    SECONDARY   SCHOOLS 


BY 

E.    R.    HEDRICK 

PROFESSOR    OF    MATHEMATICS 
THE     UNIVERSITY     OF     MISSOURI 


(^ 


NEW  YORK  •:•  CINCINNATI  •:•  CHICAGO 

AMERICAN    BOOK    COMPANY 


Copyright,  1908,  by 

E.   R.    H  ED  RICK. 

Entered  at  Stationers'  Hall,  London. 

hedbick's  algebra. 
W-  P.     I 


CIA 


€o 


MY   SISTEK    AND    TEACHER 

MRS.  ERNST  VOSS  nei:  AURIE  VAIL  HEDRICK 

^  THIS    BOOK    IS    DEDICATED 

IN    GRATEFUL    RECOGNITION    OF    HER    LONG 
DEVOTION    AND    ASSISTANCE 


I" 
I 


.'i48fi83 


PREFACE 

The  presentation  of  a  new  text  on  elementary  algebra 
to  the  public  deuiancls  justification.  It  is  the  author's 
earnest  hope  that  this  book  will  fill  a  need  felt  by  many 
teachers  for  a  book  that  is  at  once  thoroughly  modern,  yet 
conservative  of  what  was  good  in  the  older  text-books. 

The  drill  afforded  by  the  older  texts  will  not  be  found 
wanting.  The  exercises  in  all  instances  are  numerous, 
especially  in  those  topics  where  drill  work  is  needed  to 
develop  technique.  It  is  sfroii</Ii/  recommended  that  not  all 
the  examples  in  all  the  lists  be  solved  on  first  reading,  for 
the  lists  liave  been  prepared  with  a  view  to  reviews,  for 
which  some  of  the  exercises  should  be  saved. 

Omissions  from  the  text,  as  well  as  from  the  exercises, 
are  to  be  encouraged.  The  inclination  to  do  everything 
between  the  covers  of  a  book -is  one  of  the  vicious  tenden- 
cies of  the  days  when  teachers  were  drillmasters  only. 
Rather  let  wliat  is  done  be  done  with  extreme  thorough- 
ness. It  is  probably  as  valuable  a  fact  as  any  other  which 
the  student  may  learn  that  not  all  of  algebra  is  contained 
in  this  or  any  other  book  whatsoever;  he  should  be  left 
with  the  distinct  impression  that  tliere  are  many  things 
which  he  has  yet  to  learn,  —  among  them,  perhaps,  some 
of  the  topics  in  this  book. 

The  distinctive  features  of  the  book  are  matters  of 
detail.  No  one  favorite  principle  or  prejudice  has  guided 
the  author,  unless  it  be  a  desire  to  speak  the  truth,  which 
is   often    found  an  inconvenience.      The  wliole  truth  is 


vi  PKEFACE 

occasionally  withheld,  in  view  of  the  natural  limitations 
of  the  scope  of  tlie  book. 

It  has  also  been  an  aim  to  meet  the  saner  views  —  both 
radical  and  conservative  —  which  have  been  expressed  in 
recent  Reports  of  committees  throughout  the  country,  on 
the  teaching  of  elementary  algebra. 

Thus,  several  topics  traditionally  given,  and  still  required 
by  certain  colleges  for  entrance,  are  placed  in  the  Appen- 
dix rather  than  in  the  body  of  the  book.  Among  others, 
the  Euclidian  process  for  finding  the  H.  C.  F.,  the  few 
fragmentary  methods  for  solving  special  simultaneous 
quadratics,  an  explicit  treatment  of  imaginaries,  and  the 
theory  of  limits  and  infinite  series,  are  relegated  to  the 
Appendix,  in  the  belief  that  none  of  these  topics  deserve 
a  place  in  the  usual  high  school  course  except  for  special 
cause.  In  the  last  two  topics  mentioned,  the  traditional 
treatments  are  by  no  means  free  from  errors,  and  this  fact 
alone  would  indicate  that  they  are  neither  suited  to  the 
child's  intelligence,  nor  particularly  valuable  to  him. 

Extreme  rigor  of  proof  is  not  exacted,  and  the  psy- 
chological advantage  of  conviction  as  compared  with  proof 
is  recognized;  but  frank  statements  are  made  whenever 
the  proofs  are  not  complete. 

The  language  of  the  book  is  purposely  simple,  frank, 
and  conversational.  A  special  effort  is  made  to  impart  to 
tlie  student  the  ability  to  elucidate  English  in  exercises, 
and  to  translate  English  into  formulas  —  the  principal 
advantage  in  the  algebraic  notation. 

Graphical  illustrations  are  treated  as  a  normal  part  of 
;ilgel)raic  knowledge.  Tliey  are  used,  whenever  they  are 
valuable,  without  extensive  discussion  and  without  osten- 
tatious use  of  needless  nomenclature.  It  will  be  found 
tliat  no  complicated  curves  are  used  in  exercises  for  the 
student.       Teachers    who    realize   the    ti-emendous    value 


PREFACE  vii 

of  graphical  work,  botli  in  everyday  affairs  and  in  tlie 
study  of  mathematics  and  science,  will  welcome  the  con- 
sistent use  of  this  tool,  probably  the  most  useful  of  alge- 
braic tools  for  actual  problems  of  life  and  science.  The 
early  introduction  of  the  simpler  notions  will  not  alarm 
a  teacher  who  has  tried  these  ideas  Avith  very  young 
students. 

In  the  practical  exercises,  the  aim  has  been  to  eliminate 
artificial  problems  of  the  most  extreme  type  —  those  in 
which  the  data  could  not  conceivably  be  known  before 
the  answers  were  known.  Such  problems  tend  to  destroy 
interest  and  sympathy.  Problems  intended  for  drill  work 
may  be  artificial,  if  there  is  no  pretense  of  clothing  them 
in  hypocritically  practical  language. 

While  it  meets  the  entrance  requirements  of  Ainerican 
colleges  and  universities  generally,  this  book  is  written 
essentially  for  those  for  whom  the  high  school  course  is 
to  be  the  last.  Especially  for  such  students,  it  would  be 
deplorable  to  omit  the  principal  features  of  the  body 
of  the  book  in  favor  of  those  in  the  Appendix.  Such 
students  are  in  a  majority  in  most  high  schools. 

An  unusually  large  number  of  persons  have  assisted 
me  in  various  ways.  I  am  indebted  to  Mr.  W.  A.  Hur- 
witz,  for  the  preparation  of  a  large  number  of  the  exer- 
cises ;  to  Professor  L.  D.  Ames,  for  the  preparation  of 
portions  of  the  text ;  to  Professor  O.  D.  Kellogg,  Super- 
intendent J.  M.  Greenwood  (Kansas  City),  Mr.  W.  L. 
Jordan  (Des  Moines),  and  others,  for  suggestions  upon 
the  manuscript ;  to  Dr.  Louis  Ingold,  for  checking  all 
answers.  For  suggestions  upon  the  proofs,  I  woidd  espe- 
cially thank  Professor  W.  F.  Osgood  (Harvard),  Professor 
James  Pierpont  (Yale),  Professor  F.  N.  Cole  (Columbia), 
Professor  C.  A.  Waldo  (Purdue),  Professor  M.  A. 
Bussewitz  (Milwaukee  Normal  School),  Professor  D.   F. 


viii  PREFACE 

Campbell  (Armour  Institute),  Professor  E.  A.  Lyman 
(Michigan  Normal  School),  Professor  H,  C.  Harvey 
(Kirksville  Normal  School),  Professor  Ira  S.  Condit  (Iowa 
Normal  School),  Professor  A.  H.  Wilson  (Alabama  Poly- 
technic Institute),  Messrs.  E.  D,  Phillips,  E.  M.  Bainter, 
and  A.  A.  Dodd  (Kansas  City),  Mr.  A.  M.  Allison  (Sioux 
City,  la.),  Mr.  Lewis  Omer  (Oak  Park,  111.),  Mr.  R.  H. 
Jordan  (St.  Joseph).  The  book  owes  much  to  the  sug- 
gestions of  these  and  other  persons.  To  all  who  have 
assisted  rae,  I  would  acknowledge  here  my  indebtedness. 

E.   R.    HEDRICK. 


CONTENTS 

CHAPTER  PAf;B 

I.      iNTHOntTCTIOX.      §§  1-11 1 

Part  I.     Numbers  and  Signs.     §§  1-5     ....  1 

Part   II.     Preliminary   Definitions.     §§  6-11         .         .  7 

Summary 12 

11.  Mp:asurement  and  Aids  in  Expression.  §§  12-22  .  13 
Parti.     Measurement  of  Simple  Quantities;  Negative 

Numbers.     §§  12-15 13 

Part  II.     Relation  between  Two   Quantities;    Graphs. 

§§  16-22 17 

Summary 33 

III.  Addition    and    Subtuactiox;    Simple    Equations. 

§§•-^3-39 34 

Part  1.     Rules  for  Operation  ;  Parentheses.     §§  23-33  .  34 

Part  II.     Applications ;  Linear  Equations.     §§  34-39   .  55 

Summary 65 

IV.  ]\Iultiplication  and   Division;    Factoring;    Appli- 

cations.    §§  40-67 67 

Part  I.     Numbers  and  ^Monomials.     §§  40-48         .         .  67 
Part  II.     Longer  Expressions.     §§  49-55        ...  78 
Part   III.      Sijecial    ]\lultiplications ;    Factors;    Type- 
forms.     §§  56-63 .91 

Part  IV.     Applications;    English  translated  into  Alge- 
bra.    §§  64-67      .         . 103 

Summary 116 

V.     Fractions   (Common  Factors;  Reduction;  Operations; 

Proportion;  Fractional  Equations).     §§68-85  .         .  118 
Part  I.     Common   Factors;    Reduction    of   Fractions. 

§§  68-71        .         .         . 118 

Part  IL     Rules  of  Operation.     §§  72-76         .         .         .124 

Part  in.     Proportion.     §§  77-80 137 

Part  IV.     Fractional  Equations.     §§  81-85   .         .         .149 

Summary .         .         .157 

VI.     Simultaneous  Linear  Equations.     §§  86-93      .         .  159 

Summary 180 

^"IL     Simple  .Powers  AND  Roots.     §§94-107        .        .        .181 

Part  I.     Powers  and  Roots  of  Numbers.     §§  94-101       .  181 
ix 


X  CONTENTS 

CHAPTER  PAGE 

Part  II.     Powers   and   Roots    of   JMononiials.     §§  102- 

107 102 

Summary 201 

VIII.     Quadratic  Equations.     §§  108-121      .         .         .         .203 
Part  I.     Methods  of  Solution ;  Character  of  the  Roots. 

§§  108-115 203 

Part  II.     Practical  Applications;  Problems.     §116       .  216 

Part  III.  Properties  of  Quadratic  Equations.  §§117-121  223 

Summary 236 

IX.     Variation;  Indeterminate  Equations.     §§  122-126  237 

Summary 252 

X.     Simultaneous    Equations    involving    Quadratics. 

§§  127-134 253 

Part  I.     One  Linear  and  One  Quadratic.     §§  127-130  .  2.53 

Part  II.     Simultaneous  Quadratics.     §§  131-131:  .         .  267 

Summary 283 

XI.     Radicals  ;    Fractional  and  Negative  Exponents. 

§§  135-149 284 

Part  I.     Operations;     Fractional  and  Negative  Expo- 
nents.    §§  135-144 284 

Part  II.    Applications:  Radical  Equations.    §§145-149  300 

Summary          .........  312 

XII.  Equations  solved  by  Substitution.     §§  150-154       .  313 

Summary 322 

XIII.  Progressions  or  Sequences.     §§  155-160    .        .        .  323 

Part  I.     Arithmetic  Sequences.     §§  155-158           .         .  323 

Part  II.     Geometric  Sequences.     §§  159-161           .         .  328 

Summary           .........  333 

XIV.  Logarithms.     §§  162-167 334 

Summary          .........  353 

Appendix  —  Detached  Coefficients  —  Remainder  Theorem  ;  Fac- 
toring—  Choice  and  Chance;  Permutations  and  Comlii- 
nations  —  Inequalities  —  Binomial  Theorem  —  Euclidian 
Method  H.C.F.  and  L.C.M.  — Cube  Root  and  Higher  Roots 
—  Limits  and  Infinite  Series;  Irrational  Numbers  —  Imagi- 
nary and  Complex  Numbers — Simultaneous  Quadratics      .  354 

Tables .  407 

Index .  417 


ALGEBRA  FOR  SECONDARY  SCHOOLS 

CHAPTER   I.     INTRODUCTION 
PART  I.     NUMBERS  AND   SIGNS 

1.  Algebra.  The  word  algebra  is  used  as  a  convenient 
name  for  the  continuation  of  arithmetic ;  there  is  no  defi- 
nite line  between  the  two  subjects,  but  symbols  are  used 
more  freely  in  algebra  than  in  elementary  arithmetic. 

2.  Numbers.  The  first  numbers  we  learn  are  the  in- 
tegers, 1,  2,  3,  4,  etc.  The  students  know  also  another 
kind  of  numbers,  fractions,  which  are  useful  in  measuring 

quantities.     Later  we  shall  extend  these  ideas. 

3.  Signs  and  Marks.  Various  signs  and  marks,  some- 
times called  symbols,  or  characters,  are  used  as  in  arith- 
metic, and  new  ones  are  introduced  as  they  are  needed. 
A  table  of  these  is  found  at  the  end  of  this  book  (p.  407). 

Addition^  subtraction.^  multiplication,  and  division  are 
indicated  by  the  signs,  +,  — ,  X,  -^,  respectively,  as  in 
elementary  arithmetic. 

Tlius,  3  +  2,  3  -  2,  3  X  2,  3  -  2,  are  read  "  3  plus  2,"  "  3  minus  2," 
"3  multiplied  by  2,"  and  "  3  divided  by  2,"'  respectively.  The  product 
3  X  2  is  also  often  written  3  •  2,  the  dot  •  being  a  siiriplified  cross,  x  ; 
and  we  may  say  "3  times  2"  instead  of  "3  multiplied  by  2."  The 
quotient  3^2  is  sometimes  written  3:2,  also  |,  also  3/2;  the  form 
3:2  is  often  called  the  ratio  form  ;  the  form  f  (as  well  as  the  form 
3/2)  is  often  called  the  fraction  form;  but  they  all  mean  the  same 
thing-.     The  form  5  may  also  be  read  "3  over  2." 

1 


2  INTRODUCTION  [Cii.  J 

One  important  sign  is  =,  read  "  equals,"  or  "  is  equal  to." 
A  statement  that  one  thing-  is  equal  to  another  is  called 
an  equation  or  an  equality  ;  e.g.  2  +  4  =  6.  Two  num- 
bers are  equal  if,  and  only  if,  they  are  the  same  number, 
though  they  may  be  written  differently.  Thus,  2+-I:  is 
the  same  number  as  6,  but  it  is  written  differently. 

An  equality  of  two  fractions  (or  ratios)  is  called  a  pro- 
portion.     Thus,  I  =  f  (or  2:4  =  3:6)  is  a  proportion. 

The  radical  sign  V  is  used  to  indicate  square  root ; 
thus  Vl6  =  4.     For  further  uses  of  this  sign,  see  p.  9. 

4.  Abbreviations.  In  arithmetic  we  use  such  abbrevia- 
tions as  ft.  for  1  foot,  in.  for  1  inch,  hr.  for  1  hour,  etc.; 
and  we  write  36  in.  =  3  ft.,  etc.  In  algebra  we  abbreviate 
still  more.  Thus,  we  may  as  well  write  merely  /  for  one 
foot  and  i  for  one  inch,  then  36  i  =  3/.  Such  abbreviations 
often  simplify  problems  and  shorten  calculations ;  they 
make  general  statements  possible.     Thus,  instead  of  saying 

interest  =(rate)  x  (principal)  x  (time  in  years), 

we  may  say  i  =  r  x  j)  x  t 

with  the  same  advantage  that  we  gain  in  writing  Mo.  for 
Missouri ;  namely,  we  save  time  mid  space. 

There  is  nothing  mysterious  or  difficult  about  such  abbreviations  if 
they  are  once  explained.  Tt  should  be  noticed  that  the  same  letter 
may  be  used  for  different  things  in  different  problems  ;  tlius  i  was 
used  to  mean  one  inch  in  one  connection  above,  and  again  to  mean 
interest.  Such  abbreviations  are  simply  temporary,  for  a  single 
problem ;  the  student  will  not  be  confused  if  he  clearly  understands 
what  each  letter  means.  It  is  very  important,  however,  to  avoid  using 
the  same  letters  for  two  different  things  in  the  same  prohlem. 

In  using  letters  as  above,  the  sign  of  multiplication  is 
often  omitted;  thus  rpt  means  rxpxt,  and  ^ pr  means 
5  X  p  X  r;  but  it  is  not  safe  to  omit  the  sign  when  two 
facjtors  are  ordinary  Arabic  numbers,  since,  for  example, 


3-4]  JsLMBKRS   AND   SIGNS  3 

52  means  50  +  2,  not  5x2.  However,  this  may  be  done 
in  combining  numbers  and  letters,  as  in  arithmetic,  when 
we  say  that  if  in.  means  1  inch,  3  in.  means  3  inches. 

EXERCISES    I:    CHAPTER  I 

[Tables  of  symbols,  weights,  measures,  etc.,  will  be  found  at  the 
back  of  the  book.] 

1.  Express  4/4-3i  in  inches  if  /  means  one  foot  and  i 
means  one  inch. 

2.  Express  2/<  +  4  ?>i  —  14,s  in  seconds  if  li  means  one 
hour,  m  one  niiiuite,  and  *■  one  second. 

3.  If  b  denotes  one  bushel,  p  one  peck,  and  q  one  quart,  ex- 
press 3b—p  —  '2q  in  quarts.  Also  express  3  6—^^  —  2^  in 
pecks ;  in  bushels. 

4.  If  t  stands  for  one  tenth  and  h  for  one  hundredth,  ex- 
press 1  —  3  i  -|-  5  /i  in  per  cent. 

Express  67  %  in  terms  of  t  and  h. 

5.  Let  c  denote  one  cubic  centimeter,  i  one  cubic  inch,  and 
,/one  cubic  foot.  Express  yV/—  132  i  in  terms  of  c,  if  we  let 
i  equal  16  c  (which  is  very  nearly  correct). 

6.  Express  2m-\-o0y +  3f  n\  terms  of  /  if  m  =  5280/ and 

[Note  that  figures  given  are  actually  applicable  to  the  case  in  which 
/"denotes  one  foot,  y  one  yard,  and  m  one  mile.] 

V-\    7.   If  r;  =  4  q,  express  2  fj  +  3  (/  in  terms  of  q  ;  in  terms  of  rj. 

[Can  you  mention  a  case  of  measurement  to  which  this  problem 
would  be  applicable  ?] 

8.  If  y  denotes  one  yard  and  i  one  inch,  express 
9/t  =  y4-3|<:  in  terms  of  i.  What  unit  of  length  does  m 
denote  ? 

9.  If  X  denotes  a  single  thing,  d  a  dozen,  and  m  a  gross, 
express  2  m  +  11  d  —  6  ic  in  terms  of  d. 


4  INTRODUCTION  [Ch.  I 

10.  Express  a  — 17 b  +  Sc  —  oOd  in  terms  of  d  if  a  =24 6, 
&  =  60  c,  c  =  60  rf.     To  what  measures  will  this  apply  ? 

11.  What  is  the  value  of  4  a  +  3  6  —  3  c  if  a  =  2,  b  =  o,  c  =  3? 

12.  What  is  the  value  ot  6  xt  —  3  tj  -  5 1  ii  x  =  3,  y  =  0,  t  =  2? 

13.  Express  4?ii  — 3n4-10p  in  terms  of  n  if  m  =  2n, 
n  =  5 p.  What  is  the  value  of  the  result  if  n  =  52  ?  if  n  =  32  ? 
if  >i  =  0  ? 

14.  What  is  the  value  of  '^  '  '■  if  a  =  3,  &  =  4,  c  =6, 
a;  =  5,2/  =  2?                                 ^^"-""^ 

15.  li  A  =  ^a,B  =  3h,  express — in  terms  of  a  and  6. 

Ah  —  Ba 

Find  the  value  of  this  result  if  a  =  2  and  &  =  4. 

16.  If  X  denotes  the  area  of  a  square,  and  y  the  length 
of  a  side,  then  x  =  y '  y.  What  is  the  area  of  a  square 
whose  side  is  4  feet  long  ?  (The  product  y  •  y  is  often 
written  "I  is  called  the  square  of  y  because  y  -  y  is  the 
area  of  the  squ«,?e.) 

17.  Express    ^      -^    in  terms  of  x  alone  if  y=2x.     Then 

1-xy 

find  its  value  if  cc  =  ^. 

18.  What  is  the  area  of  a  rectangle  whose  length  is  a  and 
whose  breadth  is  b?  What  is  the  area  of  a  rectangle  5  inches 
long  and  3  inches  wide  ? 

19.  Let  m  denote  the  side  of  a  cube,  n  its  volume.  Express 
n  in  terms  of  m.  If  m  =  2  feet,  what  is  n  ?  If  w  =  27  cubic 
inches,  what  is  m? 

20.  Let  the  dimensions  of  a  room,  measured  in  feet,  be  de- 
noted by  ?o  (the  width),  I  (the  length),  and  h  (the  height  of 
the  ceiling);  and  let  c  be  the  cost  (in  cents)  per  square  yard 
of  plastering.  How  much  will  it  cost  to  plaster  the  walls  and 
ceiling  of  the  room,  neglecting  doors  and  windows? 

What  will  the  cost  be  if  Z  =  16  feet,  w  =  15  feet,  h  =  10  feet, 
and  c  =  15  cents? 


-5]  NUMBERS   AND   SIGNS  5 

5.  Problem  Solving.  Ahhreviations  are  especially  useful 
11  the  solution  of  problems. 

Ex.  1.  Find  a  number  such  that  the  sura  of  that  number 
uid  half  the  number  is  18. 


Souttion 

Solution 

(not  abbreviated) 

(abbreviated) 

Consider  Me  number  to  he  found. 

Let  n  - 

=  the  number  to  he  found 

Then    (that    number)  +  \    (that 

Then 

n  +  \n  =  18, 

number)  =18, 

or,  1  (that  number)  =  18, 

or, 

1  n  =  18, 

or,  \  (that  number)  =  6, 

or, 

|n  =  6, 

or.  (that  number)  =  12  (answer). 

or, 

n  =  12. 

Check: 

Check. 

12  +  1  •  12  =  18.     (Correct.) 

12 

+  |.  12  =  18.     (Correc 

A  check  is  any  process  for  testing  an  answer.  If, 
as  here,  the  answer  can  be  tried  directly  in  the  given 
problem,  the  check  is  complete  ;  a  complete  check  shows 
that  the  answer  is  surely  correct.  x\ 

[The  teacher  should  explain  carefully  the  check  on  the  answer  in 
this  problem,  and  the  students  should  be  required  to  check  their 
answers  in  all  cases.] 

Ex.  2.  A  merchant  sold  tea  for  35  cents  per  pound.  "\^Tiat 
was  the  cost  to  him  if  he  made  25  %  profit  on  the  cost? 

Solution  Solution 

(not  abbreviated)  (abbreviated) 

Consider  the  cost  to  merchant.  Let  c  =  the  cost  to  merchant. 

Then  the  profit  to  merchant  is  25%  Then  profit  to  merchant  =  25  %  (^- 

of  cost  to  merchant, 

or  \  of  cost  to  merchant.  Then  profit  to  merchant  =  \c. 

Hence,  the  total  price  is  the  cost  Hence  total  j^rice  is  c  +  ^  e. 

to  merchant  added  to  \  of  cost 

to  merchant, 

or  J  of  cost  to  merchant.  Hence  total  price  =\c. 

Then  35  cents  is  |  of  cost  to  mer-  Then  35  cents  =  |  c, 

chant, 
or  28  cents  is  the  cost  to  merchant.  28  cents  =  c. 

Check:   28  +  i  .  28  =  35.  Check:         28  +  i  •  28  =  35. 


6  INTRODUCTION 

EXERCISES   II:    CHAPTER   I 

[Check  each  result  as  indicated  in  the  preceding  examples.] 

1.  If  the  sum  of  three  times  a  certain  number  and  half  that 
number  is  14,  what  is  the  number  ? 

[Let  the  students  invent  and  propose  to  each  other  such  pi-oblems.] 

2.  What  number  yields  a  remainder  5  if  diminished  by 
the  sum  of  one  half  of  itself  and  one  third  of  itself  ? 

3.  What  must  be  the  amount  of  money  invested  in  an  enter- 
prise yielding  2;")  %  profit,  in  order  that  the  money  in  baud  at 
its  conclusion  may  be  f  1000  ? 

4.  A  man  attempted  to  charge  8  %  interest  on  money. 
Being  able  to  collect  only  the  legal  rate  (6  %)  he  made  $20  less 
than  he  expected.     What  was  the  amount  of  money  loaned? 

5.  A  merchant  buys  butter  for  30  ^  a  pound  and  sells  it 
for  3G  i  a  pound.     What  is  his  per  cent  of  profit  on  the  cost  ? 

6.  A  fruit  dealer  sold  oranges  at  5  cents  apiece  or  50  cents 
per  dozen.  He  found  that  he  received  $54.00  for  100  dozen. 
How  many  were  sold  singly  and  how  many  by  the  dozen  ? 

7.  What  must  be  the  per  cent  of  profit  on  an  investment  if 
$525  is  to  produce  $600? 

8.  A  shoe  dealer  buys  100  pairs  of  shoes  at  $  2.00  each  and 
sells  75  pairs  at  $2.50  each.  To  sell  the  remaining  25  pairs 
he  marks  them  down  so  as  to  make  20%  profit  on  the  whole. 
What  is  the  price  per  pair  that  will  give  this  result? 

9.  I  propose  to  a  friend  the  following  puzzle  :  "  Think  of 
any  number,  add  5  to  it,  multiply  the  result  by  3,  and  subtract 
4."     He  gives  the  result  as  17.     What  was  his  number  ? 

10.  I  have  an  opportunity  to  lend  money  at  5%  simple 
interest  for  a  period  of  8  years.  How  much  must  I  lend  in 
order  that  the  amount  may  be  $3500? 

11.  How  large  a  cubical  box  may  be  covered  with  24  square 
inches  of  paper  ? 


PART    II.     PRELIMINARY    DEFINITIONS 

6.  Expressions.  All  single  groups  of  numbers  and  char- 
acters of  the  kind  already  used  are  called  expressions ;  in 
order  to  deserve  this  name,  the  group  of  numbers  and 
characters  must  have  some  meaning. 

Thus,  8/  +  2  i  has  a  meaning  if  /  means  1  ft.  and  i  means  1  in. ; 
then  3/'+  2i  is  an  expression. 

Again,  the  group  of  characters  p  +  rpt  is  an  expression,  although 
the  meaning  is  not  wholly  clear.  It  is  at  least  clear  that  some  num- 
ber p  is  to  be  increased  by  the  product  of  three  quantities,  r,  p,  and  /. 
The  meaning  becomes  clearer  if  we  are  told  thatyj  means  the  principal, 
/•  the  rate,  and  t  the  time  in  years  in  a  certain  interest  problem;  then 
the  above  expression  clearly  means  the  amount.  The  meaning  be- 
comes still  clearer  when  the  numerical  values  of  r,  p,  and  t  are  given, 
say  r  =  5%,  p  =  ^  125,  and  I  =  3  years.  These  various  stages  in  clear- 
ness do  not  contradict  the  fact  thatyj  -|-  rpt  has  in  itself  (without  any 
further  explanation)  a  certain  meaning  as  expressed  above. 

7.  Terms.  An  expression  may  contain  one  or  more  + 
or  —  signs,  which  separate  it  into  parts  ;  these  parts  are 
called  the  terms  of  the  expression. 

Thu-s  in  p  +  rpt  the  terms  are  /;  and  rpt.  In  2  ax  -f     ■  _    "  -(-4  c  the 

terms  are  2  ax  and      '^      ~  and  A:C.    To  be  sure,  the  term  '''  ■'      "  itself 

3  3 

contains  a  +  sign,  but  it  is  not  separated  into  parts  by  that  sign  as  it 
stands.  On  the  whole,  the  word  term  is  used  rather  loosely,  the  inten- 
tion being  to  distinguish  those  parts  which  stand  in  rather  compact 
groups  as  compared  with  the  rest. 

The  terms  of  an  expression  are  calculated  separately,  and  these 
results  are  added  or  subtracted  in  their  order.  Thus,  as  a  general 
rule,  the  multiplications  and  divisions  are  carried  out  first,  after 
which  the  terms  thus  formed  are  added  or  subtivacted.  If  anything 
else  is  intended,  parentheses  are  used  to  show  that  the  expression  inside 
the  parentheses  is  to  be  calculated  first.  Thus,  the  expression 
2  ax  -f  m  (") ;/  +  2)  -(-  4  c  has  three  terms :  2  ax,  m  (.5  y  +  2),  and  4  c ; 
the  term  m  (5  ij  -f  2)  means  m  times  the  sum  of  5  ^  and  2.  See  also 
§11,  p.  10. 

7 


8  INTRODUCTION  [Uh.  I 

To  distinguish  expressions  the  special  names  monomial, 
binomial,  and  trinomial  are  often  used  for  expressions 
that  liave  one  term^  two  terms,  or  three  terms,  respectively. 

Thus  rpt  and      -^^     "   are  monomials,  p  +  rpt  is  a  binomial,  and 

5  w  +  '•* 
2  ax  +  '  ■■'  —  4  c  is  a  trinomial,  etc. 

3 
When  we  wish  to  refer  to  a  complicated  expression  of  more  than 
three  terms,  we  may  simply  call  it  an  expression.     The  word  polyno' 
mud  may  also  be  used  in  certain  simple  cases  defined  in  §  9  below. 

8.  Factors.  When  several  numbers  or  expressions  are 
multiplied  together,  any  one  of  them  is  called  a  factor  and 
the  result  is  called  the  product.v 

Thus  the  factors  of  the  product  rpt  are  r,  p,  and  t. 

Any  factor  is  called  the  coefficient  of  the  rest  of  the 
product ;  usually,  however,  the  word  eoeffiaierit  is  under- 
stood to  be  that  factor  which  is  represented  by  Arabic 
numerals,  or  which  is  supposed  to  be  a  known  number,  un- 
less the  coefficient  of  a  special  part  is  required. 

Thus,  in  2  axy  the  coefficient  is  2  unless  a  specific  part  is  mentioned, 
but  if  we  are  asked  for  the  coefficient  of  x//,  the  answer  is  2  a ;  the  co- 
efficient of  2  ax  is  y,  and  so  on.  Coefficient  is  but  another  name  for 
multiplier.  Thus,  in  2  axy,  2  is  the  multiplier  of  axy  and  2  a  is  the 
miiltiplier  of  xy ;  but  coefficient  is  the  word  generally  used. 

If  the  coeflBcient  is  1,  it  is  not  written,  since  multiplying  a  number 
by  1  does  not  change  it.  Thus,  in  axy  the  coefficieiit  is  1  unless  the 
coefficient  of  a  special  part  is  required. 

Terms  that  are  precisely  the  same  or  that  differ  only  in 
their  coefficients  are  called  similar  terms  or  like  terms. 
'I'lius,  ''Ix  and  3  a:;  are  similar;  3  r/Ai  and  A  nfl)i  arc  similar. 
Terms  that  are  not  similar  are  called  dissimilar  or  unlike. 

9.  Powers.  The  product  of  two  e(|ual  factors  is  called 
tlie  square  of  tliat  factor ;  it  is  indicated  by  a  small  figure 
2  at  the  upper  right-hand  corner.     Thus,  x-  x  =  x^. 

The  product  of  three  equal  factors  is  called  the  cube  of 
that  factor.     Thus,  x  -  x  '  x  =  ofi. 


7-10]  PRELIMINARY   DEFINITIONS  9 

The  product  of  any  number  of  equal  factors  is  called  a 
positive  integral  power  or  a  simple  power  of  that  factor ; 
it  is  indicated  by  writing  the  number  of  the  factors  at  the 
upper  right-hand  corner  of  the  factor. 

Thus,  we  write  x  ■  x-  X'X  =  x'^ -,     X'  x-  x- x  •  T  =  r';  etc. 

The  number  of  factors  is  called  the  exponent  of  the 
power.  Thus,  in  a;*  the  exponent  is  4,  and  x^  is  called  the 
fourth  power  of  x,  or  simply  x  fourth  potver.  We  shall  later 
extend  these  definitions  of  power  and  exponent  so  as  to  give 
a  meaning  to  such  forms  as  a;%  x~^,  aP,  etc.,  which  are  at 
l^resent  meaningless  to  the  student.     See  pp.  286,  296. 

Since  any  number  x  may  be  considered  as  taken  once  as 
a  factor  to  make  itself,  x^  means  the  same  as  x.  Hence,  it 
is  useless  to  write  the  exponent  1,  and  when  no  exponent 
is  written,  1  is  understood. 

A  polynomial  in  certain  given  letters  is  an  expression 
whose  terms,  if  they  contain  one  of  those  letters  at  all,  con- 
tain it  as  a  factor  which  is  a  simple  power.* 

10.  Roots.  If  a  number  is  the  product  of  two  equal 
factors,  one  of  the  equal  factors  is  called  the  square  root 
of  the  number;  if  a  number  is  the  product  of  three  equal 
factoVs,  one  of  the  equal  factors  is  called  the  cube  root. 
In  general,  if  a  quantity  is  the  product  of  a  number  of  equal 
factors,  one  of  them  is  called  a  root  of  that  quantity  ;  and 
the  number  of  the  factors  is  called  the  index  of  the  root. 

Thus,  4  has  two  equal  factors,  2  and  2.  Heuce,  the  square  root  of 
4  is  2,  written  Vi  =  2  ;  likewise,  since  3  •  3  •  3  =  27,  the  cube  root,  of 
27  is  3,  written  V27  =  3.  Again,  since  a  •  a  =  a^,  Vcfi  =  a.  So  also, 
Va^  =  a ;  a/c^*  =  a,  etc.  The  index  is  always  written  at  the  upper 
left-hand  corner  of  the  sign  V  ,  except  for  square  roots,  which  are  in- 
dicated by  V    instead  of  v  .      These  ideas  will  be  extended  later. 

*A  broadt-r  use  of  polynomial  is  common  in  many  text-books,  but  it  is 
not  good  usage.  Standard  mathematical  works  otlier  than  text-books 
agree  on  the  detiiiitinii  used  above.  The  word  is  not  used  extensively  in 
this  book  until  alter  p.  88,  where  a  more  detailed  explanation  is  given. 


10  IXTKODUCTION  [Ch.  1 

11.  Parentheses.  When  we  wish  to  group  terms  to- 
gether, we  use  ordinary  parentheses  (see  §  7,  p.  7).  Thus, 
(^a  +  b)^  means  the  square  of  the  expression  a  +  b. 

In  order  to  avoid  confusion  wlien  one  pair  of  paren- 
theses occurs  within  another,  we  use  different  forms,  as 
follows:    [  ],  called  brackets;  \  j,  called  braces;  and  , 

called  the  vinculum.  These  names  are  convenient  to  dis- 
tinguish the  shapes,  but  they  are  all  used  alike,  and  they 
are  all  called  parentheses  except  when  some  confusion 
would  result. 

Thus  (2  +  3)  •  4  means  5  •  4,  or  20;  {n -\-  h)c  means  the  sum  of  a  and 
b  multiplied  by  c;  2Va  +  b  means  twice  the  square  root  of  the  sum 
(«  +  /»).  We  may  now  use  more  complicated  expressions,  the  con- 
venience of  which  will  be  seen  in  the  next  chapter.     For  instance, 

2(a+6)(c  +  fZ)  +  (a+i)2 

means  t^vice  the  product  of  (a  plus  h)  and  (r  plus  d)  added  to  the 
square  of  (a  plus  h).  If  the  letters  used  mean  certain  numbers,  say 
a  =  2,  ft  =  1,  c  =  3,  fZ  =  4,  then  a  +  ft  =  3,  c  +  rf  =  7,  and 

2(a  +  ft)(c  +  fl)  +  (n  +  ft)2  =  2.3-7  +  3^  =  42  +  9  =  51. 

EXERCISES    III  :    CHAPTER   I 

Find  the  value  of : 

1.  2  a^  if  o  =  2 ;  if  a  =  3  ;  if  a  =  5. 

2.  4  a^&3  if  ^,  ^  2,  /j  =  3 ;  if  a  =  1,  />  =  2  ;  if  a  =  5,  6  ==  ? . 

3.  V;^-  if  p  =  2,  r  =  8  ;  if  p  =  3,  r  =  12  ;  iip  =  18,  r  ^  8. 

4.  x"'  +  4:  —  4:X  —  X-  \i  x  =  l;  if  ;«  =  2  ;   if  x  =  3  ;   if  x  =  4. 


5.-V5rt-W'  +  4a-((!>  +  c)    if    a  =  o,    b  =  2,    c  =  o;    if   ci  =  20, 
6  =  1,  0  =  2.  

6.   Find   the  value  of   ^' Vl -/-?/ Vl  -  ^-   if  ^  _  .<i^  ,/^  | 
xy  -\-  Vl  —  a^  •  VI  —  y'' 


^^    j,V^^y^  +  yVa?-l^  ifa  =  5,  ;>  =  3,.x-  =  13.:v  =  5. 
^/{x'-y%a'-b')-by 


11]  rRELIMIXAllY    DEFINITIONS  11 

8.  P>"if  v=2;  if  n  =  8;  if  ?(  =  1. 

9.  X"  iix=-2,    y  =  o;    if    a;  =  4,    y  =  l;    if   .r=l,    y  =  2. 

10.  -^vTtif  a  =  64,?>  =  a,c  =  2;  if  a  =  256,  ?.  =  4,  c=  2. 

Express : 

11.  (2/+  3  /)-  in  square  inches  if  /=  one  foot  aud  /  =  one 
inch. 

12.  (2  (  +  u)-  in  units  if  t  denotes  one  ten  and  u  one  unit. 
Then  write  the  result  in  terms  of  t'-,  t,  and  x. 

Show  that : 

13.  (x  +  y)-  =  x'  +  2xy  +  y-  if  a;  =  3,  y  =  2;  if  x  =  5,  t/  =  l; 
if  x^h,y  =  h 

14.  (,,2  ^  ly  ^  („2  _  1)2  ^  (2  vy  if  n  =  1 ;  if  n  =  2 ;  if  ?i  =  3 ; 
if  n  =  4. 

15    '!!±A'  =  a-  -  o^>  +  h'  if  o  =  3,  ?>  =  2 ;  iia  =  5,h  =  1. 

16.  a;-— (r  +  s)j-+?-s  =  (a;-7-)(.'B  — s)    if   x  =  4:,  r  =  l,  .s  =  2; 
if  X  =  I,  r  =  I,  s  =  1 ;  if  a;  =  3,  r  =  3,  s  =  1. 

17.  (p^  +  9^)(p  +  g)(i> - q)=p' -q'  iip--=2,  q  =  l;  if  p  =  3, 
q  =  2. 

18.  In   Ex.    1    name   the   coefficient   of    a^';    the    exponent 
of  a. 

19.  In  Ex.  2  name  the  coefficient  of  a-b^;  of  cr;  of  1/;  the 
exponent  of  (f ;  of  b. 

20.  In    Ex.   4   name   the   coefficient   and   the   exponent   in 
each  term. 

21.  In  Ex.  5   what  is   the  coefficient  of  «-   in   the  second 
term  ? 

22.  Name   the    exponent   of   3   in    Ex.   8;    of  x  in  Ex.   9. 
Name  the  coefficients  in  the  result  of  Ex.  12. 

23.  Give  some  factors  of  the  expressions  found  in  Ex.  1 ; 
in  Ex.  2. 


12  INTRODUCTION 


SUMMARY    OF    CHAPTER    I:     INTRODUCTION    TO    ALGEBRA 

pp. 1-11 

Part  I.     Numbers  and  Signs.  pp-  1-6. 

Algebra:  continuation  of  arithmetic.  §  1,  P-  1- 

Known  Numbers:  integers  and  fractions.  §  2,  p.  1. 

Siyna  and  Marks :  known  symbols;  x,  ^,  +,  — ,  with  variations; 
V  ,  meaning-  square  root;    =,  meaning  equality.      §  3,  pp.  1-2. 

Use  of  Letters  for  Numbers  ;  Abbreriatlons  :  illustration,  i  =  r  -p  -t; 
caution  against  use  of  same  letter  for  different  things.  Exer- 
cises I.  §  4,  PP-  2-5. 

Pruhlem  Solving:  comparison  of  arithmetic  and  algebraic  solutions; 
check  on  answers.     Exercises  II.  §  5,  pp.  5-6. 

Part  II.     Preliminary  Definitions  of  Certain  Words. 

pp.  7-11. 

Expression :  group  of  symbols  with  meaning.  §  6,  p.  7. 

Term:  part  of  an  expression  set  off  by  +  or  —  signs;  monomial, 
one  term ;  binomial,  two  terms ;  trinomial,  three  terms ;  ex- 
pression, any  number  of  terms.  §  7,  pp.  7-8. 

Factors :  nuxltiplication  gives  product. 

Coefficient:  a  multiplier,  usually  the  Arabic  factor. 

Similar  Terms  :  identical  except  coefficient.  §  8,  p.  8. 

Simple  Power:  product  of  identical  factors. 

Exponent:  the  number  of  these  equal  factors.  §  9,  pp.  8-9. 

Polynomial :  sum  of  simple  terms,  only  simple  powei's. 

Roots:  reverse  of  powers.  §  10,  p.  9. 

Parentheses :  group  terms  into  one,  see  also  §  7,  p.  7.  Exercises 
III,  algebraic  notation.  §  11,  pp.  10-11. 


CHAPTER    II.     MEASUREMENT     AND    AIDS    IN 
EXPRESSION 


PART    I.      MEASUREMENT     OF     SIMPLE 
NEGATIVE   NUMBERS 


QUANTITIES. 


12.    Simple    Measurement.      In    measuring    lengths   we 
are  used  to  marking  off,  on  a  straight  line,  equal  distances 
wliich  represent  equal  units,  as  q       i       2       3      4       5       e 
in  Fig.  1.    It  is  often  convenient   I       I        I        I        I        1        l~ 
to  sub  divide  these  units,  eitlier  ^'^-  ^■ 

into  tenths  or  into  some  otlier  number  of  parts.  Fig.  2 
represents  part  of  a  ruler  divided  into  inches  and  eighths 
of  an  inch. 


Fig.  2. 


Fig.  3  represents  part  of  a  meter  stick  ruled  in  centi- 
meters and  millimeters. 


20 


Fig.  3. 

13.  Thermometers.  Otlier  kinds  of  quantity  are  often 
measured  similarly.  Thus,  a  thermometer  is  marked  off 
in  divisions,  each  of  which  corresponds  to  one  degree. 

The  starting  point,  which  we  call  zero,  on  a  Fahrenheit  thermome- 
ter, was  put  where  it  is  because  the  inventor  could  not  artificially 
cause  a  lower  temperature  by  means  of  his  snow  and  salt  mixtures. 

13 


14  MEASUREMENT   AND   EXPRESSION  [Cn.  11 

14.  Negative  Numbers.  The  thermometer  may  fall 
even  heloiv  zero ;  it"  it  does,  we  ordinarily  say  that  the 
temperature  is  so  many  degrees  "  below  zero." 

There  is  an  abbreviated  way  of  saying  the  same  thing. 
Let  us  suppose  the  temperature  is  60°  at  noon  and  falls  20° 
by  night.  Then  at  night  the  temperature  is  60°  —  20°  =  40°. 
But  if  it  is  0°  at  noon  and  falls  20°  by  night,  tlien  at 
night  we  say  it  is  0°—  20°,  thereby  meaning  20°  heJoiv  zero. 
Thus,  "  20°  heloiv  zero  "  is  often  written  simply  "  —  20°," 
the  0°  being  omitted  from  the  expression  0°  -  20°  because 
it  is  useless. 

This  system  works  very  well.  For  example,  if  it  was  30°  above 
zero  and  has  fallen  40°,  meanwhile  it  must  have  been  exactly  zero  at 
some  time,  just  after  it  had  fallen  30°.  It  then  must  have  fallen  10° 
more,  making  10°  helow  zero  or  -10°.  That  is,  30°  -  40°  =  - 10°. 
This  result  is  very  convenient,  for  30°- 40°  ought,  as  above,  to  be 
equal  to  30°-  30°-  10°  or  0°-  10°  or  - 10°. 

This  abbreviation  consists,  then,  in  writing  ''  —  10°  "  for 
"10°  helow  zero^'  etc.;  in  general  —  n°  —  n°  helow  zero. 

Similar  quantities  arise  frequently.  For  example,  if  a  man  has  $40 
and  spends  !|15,  we  say  he  has  |40  -  .f  15,  or  |2."i.  But  if  he  has  f  40 
and  spends  $.50,  we  say  he  is  in  debt  ^10.  It  is  simpler  to  express  this 
as  "  --110." 

Again,  if  a  weight  of  20  pounds  and  a  weight  of  30  pounds  are 
hung  together  on  a  scale,  the  total  weight  is  20  pounds  +  30  pounds 
=  50  pounds;  but  if  a  weight  of  50  pounds  is  attached  to  a  balloon, 
the  total  weight  is  diminished.  We  therefore  say  the  balloon  weighs 
less  than  zero  and  indicate  this  by  writing  the  minus  sign  in  front  of 
the  amount  it  pulls  up.  Thus,  if  a  .50-pound  weight  is  reduced  to  30 
pounds  by  tying  a  balloon  to  it,  we  say  the  balloon  ''pulls  up  by  an 
amount  of  20  pounds  "  ;  or  that  "  its  iceight  is  -20  pounds." 

Numerous  examples  of  this  sort  may  be  given,  wi^h 
pairs  of  quantities  tliat  are  exactly  the  opposites  of  each 
other,  so  tliat  ecpial  amounts  of  the  two  kinds  counter- 
balance or  destroy  each  other.  Thus  mone//  itud  debts, 
iveiijlits  and  balloons^  temperatures  above  and  below  zero, 


Hi:.]  SIMPLE    (H'ANTlTIJvS  15 

and  so  on,  are  such  pairs.  One  being  chosen  as  tlie 
original  quantit}'  considered,  the  contrasting  one  is  denoted 
by  the  sign  —  prefixed  to  its  amount.  The  contrasting 
one  (denoted  by  the  sign  —  )  is  calhjd  negative,  the  original 
one  being  called  positive. 

This  arrangement  can  in  any  case  be  reversed.  Thus  iiionej^  'ii^y 
be  thought  of  as  the  opposite  of  a  debt;  then  to  /<«re  !i?10=  -  ^  10  in 
debt,  but  we  always  try  to  choose  the  simpler  way  in  any  given  case. 
Sometimes  it  is  quite  immaterial  which  quantity  is  called  positiiy 
and  which  is  called  negative.  Thus,  walking  east  may  be  regarded 
as  positive  progress  if  one  really  wants  to  go  east.  If  a  man 
who  is  lost  wanders  10  miles  west  when  he  intended  to  go 
east,  we  may  say  he  has  gone  —  10  miles  east.  In  this  case,  ^' 
the  circumstances  determine  whether  going  east  or  going 
s\'est  is  called  positive ;  but  if  one  is  called  positive,  its  g. 
opposite  thereby  becomes  negative.  2 

It  is  important  to  keep  the  same  agreement  throughout  any       l- 
one  problem,  after  a  choice  has  been  made.  ^ 


-1 

-3  - 

-4 


15.  Representation  of  Negatives.  We  may  always 
represent  negative  quantities  just  as  on  a  ther- 
mometer. Drawing  a  vertical  straight  line,  we  -5- 
shall  agree  that  upwards  is  positive,  then  down-  ~®"" 
wards  is  negative.  The  starting  point  is  marked 
zero.  Successive  steps  upward  are  marked  1,  2,  *'"^-  *• 
3,  4,  5,  and  so  on,  the  distances  between  marks  being 
equal.  The  similar  equal  spaces  below  the  starting  point 
are  marked  —  1,  —  2,  —  3,  etc.,  as  on  a  thermometer. 
The  number  zero  mentioned  above  is  very  important. 

It  is  easy  to  compute  on  this  scale.  If  we  wish  to  add  1  to  2,  for 
example,  we  start  at  2  and  take  4  upw^ard  steps,  thus  reaching  0. 
If  we  wish  to  subtract  3  from  7,  we  start  at  7  and  take  3  steps 
down,  thus  reaching  4.  But  we  may  perform  other  operations, 
some  of  which  might  seem  difficult.  Thus  we  may  add  5  to  —  2 
by  starting  at  —  2  and  taking  5  steps  upward,  thus  reaching  3.  Or 
we  may  subtract  5  from  3  by  starting  at  3  and  taking  5  steps 
downward,  thus  reaching  —  2.  These  operations  ai-e  new  to  the 
student. 


16  MEASUREMENT   AND   EXPRESSION 

The  line  used  need  not  be  in  any  special  position  ;  often 
it  is  drawn  horizontal,  and  distances  to  the  riglit  are  then 
usually  called  positive ;   to  the  left,  negative. 

-3  -3  -1  0+1+2  +3  +4 

liiiiiiiiili mill  iiiiililiiiil !  1  iiliiiiiiiiiliiliiMiillMiiiiiil 

Fig.  5., 

Fractions  are  indicated,  as  on  a  ruler,  by  subdividing 
the  unit  steps.  Thus  2^  is  the  point  halfway  between  2 
and  3;  —  2i  is  the  point  halfway  between  —2  and  —3; 
and  so  on,  in  a  natural  manner.  A  convenient  division 
of  the  unit  is  into  tenths,  as  indicated  in  the  figure.  We 
shall  discuss  this  furthei'  in  Chapter  III. 

EXERCISES   I:  CHAPTER   II 

Add :  Subtract : 

1.  5  to  -  1.  5.    3  from  1. 

2.  3  to  -  4.  6.   2  from  -  2. 

3.  2  to  —  2.  7.    4  from  4. 

4.  7  to  —  3.  8.    7  from  2. 

Locate  on  a  line  the  points  that  correspond  to  the  following 
numbers : 

13.    -4.3. 


9.   31.       10.   2f 

11. 

-H- 

12.    61. 

Add : 

14.   31  to  4. 

16. 

4  to  -  2.4. 

15.   3Lto4.2. 

17. 

2.3  to  -  4.2, 

Subtract : 

18.    r>  from  3.2. 

20. 

4.2  from  3f. 

19.    3  from  -  li 

21. 

1.5  from  —  1 

PART    II.     RELATIONS    BETWEEN   TWO    QUANTITIES; 

GRAPHS 

16.  Representation  of  Quantities.  Let  iis  suppose  that  a 
thermometer  is  read  every  hour,  and  that  its  reading  is  put 
down.     We  shall  have  a  table  similar  to  that  which  follows  : 


A.M. 

NOO.N 

P.M. 

Time  of  Day 

8 

9 

10 

11 

12 

1 

2 

3 

4 

5 

6 

7 

etc. 

Temperature 

46^' 

48° 

51° 

55° 

50° 

57° 

.56° 

52° 

50° 

45° 

4.3° 

42° 

etc. 

These  numbers  may  be  represented  on  a  double  scale 
on  paper  marked  with  lines  both  ways,  i.e.  horizontally 
and  vertically.  The  diagram 
represents  such  a  figure. 

The  time  i,s  measured  in  this  fig- 
ure along  the  liorizontal  line,  and 
tlie  various  hours  are  marked  at 
equal  spaces,  a.m.  being  at  the  left 
and  P.M.  at  the  right  of  the  start- 
ing point.  The  temperatures  are 
measured  in  the  figure  vertically  as 
marked.  At  8  a.m.  the  temperature 
is  40°.  We  indicate  this  by  a  small 
cross  just  above  "  8  a.m."  and 
opposite  "  40° "  in  the  figure.  At 
9  A.M.  the  temperature  is  48°;  this 
is  indicated  by  the  cross  above 
9  A.M.  and  opposite  48°.  The  various 
marks  indicate  the  various  tempera- 
tures at  the  different  hours  of  the  day  as  given  in   the  table. 

In  this  figure  we  have  represented  positive  temperatures  only, 
i.e.  temperatures  "above  zero."  If  we  take  some  day  on  wliich  the 
temperature  "  falls  below  zero,"  we  must  use  the  part  of  the  vertical 
line  marked  with  minus  signs,  as  shown  in  the  figures  on  pp.  18 
and  19. 

17 


,      :. 

:  :  :-i._::^__^"_,:         _  -:__ 

.  .-  _  -Kiiii-:-..;              -  - 

-  -8-  -9-  lull--  -1-  -2  ^  -4-  -§-  -6  -7'  8  -9  - 

|_                       -■ ..        - 

Fig.  6. 


18  MEASUREMENT   AND   EXPRESSION 

On  a  certain  day  the  temperatures  were: 


[Cu.  II 


Time  of  Day 

8 

9 

10 

0 

11 

4 

12 
10 

1 

2     3 
20    18 

4 

k; 

5 
15 

6 
12 

7 
10 

5 

i) 
_  2 

Temperature 

-10 

-  {', 

The  diagram  that  corresponds  to  the  preceding  table  is  shown  in 


Fio-. 


C\^ 

* 

5 

nS 

, 

)C 

* 

0" 

T 

mt 

f 

i 

) 

I 

6 

1 

1 

1 

2 

1 

^ 

3 

4 

I 

I 

( 

5 

8 

)'' 

1 

0 

» 

-10" 

8 

^ 

Fig.  7. 


If  the  temperature  is  read  oftener,  say  every  fifteen  minutes,  a 
much  better  figure  results.     Such  a  figure  is  shown  in  Fig.  8. 


t 

~" 

. 

' 

• 

• 

• 

, 

• 

• 

• 

• 

• 

• 

• 

• 

0 

Tiinf 

i 

a 

1 

0 

1 

1 

1 

2 

1 

2 

3 

4 

i 

i 

^ 

) 

• 

»■ 

• 

3 

• 

- 

0 

a, 

S 

^ 

. 

_.. 

i-'IU. «. 


ir,] 


RELATIONS  —  GRAPHS 


19 


Let  the  student  judge  as  nearly  as  he  can  from  this  figure  the 
temperature  at  8.:)()  a.m.,  at  11. 1.5  a.m.,  at  2  p.m.,  at  9.15  p.m.  Notice 
tliat  this  cannot  be  told  accurately,  but  only  approximately.  Notice 
also  that  the  temperature  can  be  estimated  at  times  when  it  was  not 
really  measui-ed.  Thus  the  temperature  at  11.40  a.m.,  though  not  actu- 
ally measured,  may  be  judged  by  the  figure  to  have  been  about  8°. 
A  smooth  curve  drawn  through  the  points,  as  shown  in  Fig.  9,  gives 
a  very  accurate  idea  of  what  the  temperature  was  at  any  moment. 


— 

— 

!    1 

' 

f 

V 

s 

) 

., 

■••' 

V 

/ 

*« 

^ 

/ 

r* 

■N 

X 

'^ 

> 

f 

s, 

/ 

^ 

/ 

V 

/ 

d" 

T 

>n< 

> 

L_ 

8 

J 

J 

0 

1 

1 

1 

2 

1 

3 

5 

J 

:5 

S-» 

r 

s 

/ 

V 

/ 

> 

/ 

-10 

5, 

_ 

_ 

_ 

L 

L 

L 

_ 



Fig.  !I. 

T]ie  figure  may  be  used  also  in  another  way.  We  may 
ask,  "At  what  time  was  the  temperature  14°?"  By  fol- 
h)wing  the  horizontal  line  through  the  mark  "  14  "  till  it 
strikes  the  curve,  we  see  that  the  temperature  was  14°  at 
about  1.20  P.M.  and  again  at  about  5.15  p.m.  It  should 
be  noticed  that  there  are  two  answers. 

Let  the  student  select  the  time  of  day  from  the  figure  when  the 
temperature  was  18^;  10°;  0°.  Notice  that  it  was  never  30°  on  that 
day.  Let  the  student  actually  use  a  thermometer  and  draw  such  a 
figure  as  that  just  given,  and  let  him  tiien  show  what  the  temperature 
must  have  been  (approximately)  at  some  time  vdien  he  did  not  meas- 
ure it.    Let  him  also  show  (about)  when  the  temperature  was  (say)  40°. 

Such  figures  as  those  just  drawn  are  often  called  diagrams, 
or  graphs.  Drawing  such  a  diagram  or  graph  is  called 
plotting  it  ;   locating  a  point  is  called  plotting  the  point. 


20 


MEASUREMENT   AND   EXPRESSION  [Ch.  II 


17.    Scales  of  Prices.     Many  quantities  may  be  drawn 
in  figures  like  tlie  preceding.      Following  are  examples : 

Ex.  1.     If  butter  is  30  cents  per  pound,  draw  a  figure  to 
represent  the  cost  of  any  number  of  pounds. 
We  first  make  a  table  as  shown  below : 


No.  of  lb. 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

etc. 

Price  in  f- 

30 

(50 

'.10 

120 

150 

180 

[Let  the  student  fill  in  the  blank  spaces.] 

Then,  representing  the  number 
of  pounds  of  butter  on  a  horizontal 
line,  as  shown,  and  the  cost  on  a 
vertical  line,  as  shown,  we  have  as 
before  a  set  of  points,  A,  B,  C,  D, 
E,  ••-.  Joining  these  by  a  smooth 
curve,  we  get  a  close  idea  of  the 
cost  of  any  number  of  pounds,  even 
a  fractional  number.  For  example, 
the  cost  of  3^  pounds  is  f  1.0.5,  as 
the  figure  indicates.  The  "curve" 
in  this  case  is  really  a  straight  line  ; 
this  may  happen  in  any  example, 
but  the  word  curve  is  used  at  least 
until  we  are  sure  the  line  is  really 
straight.  Most  price  curves  are 
straight  lines.  See  §§  18,  19,  pp. 
23,  21. 

Notice  that  no  butter  cost  no 
money.  Thus  at  the  point  marked 
"0"  on  the  horizontal  line  vie 
should  draw  the  point  at  the  he!(/ht  0. 
Notice  that  hiii/iiu;  —  2  pounds  of 
hitter  means  selling  2  pounds.  The 
co.'it  of  —  2  pounds  is,  therefore,  less 
than  nothing;  in  fact,  a  man  is,  of 
courae,  paid  for  the  butter  if  he  really  sells  it.  Instead  of  saying  that 
he  is  paid  60  cents  for  sellinfj  2  pounds,  we  may  say  that  the  cost 
is  _  60  cents  if  he  buys  -  2  pounds.     This  would  be  represented  in 


Cost 

200  ^'=$2 

_/ 

7 

L 

1 

_      7 

r 

JE 

_    7 

T 

-4S 

100  (Z'=$l 

.7 

T 

M 

J 

r 

~lf 

50  c'        /I 

3 

T 

r 

/\A 

10  g'  7 

^t 

Zf). 

1  0* 

•?* 

7     : 

t     - 

J 

7 

r 

Fig.  1(1. 


17] 


RELATIONS—  GRAPHS 


21 


the  figure  by  a  point  corresponding  to  —  2  on  the  horizontal  line  and 
to  —6(1^-  on  the  vertical  line.  With  this  understanding  the  curve  of 
prices  just  above  is  an  unbroken  straight  line. 

EXERCISES    II  :     CHAPTER   II 

1.   Draw  a  figure  to  represent  the  temperature  at  various 


times  of  the  d 

ay  from  the  following 

table 

A.M. 

M. 

P.M. 

Time     .     ,     . 

8 

9   |lO 

11 

12 

1 

2 

3 

4 

5 

6 

7 

8 

9  etc. 

Temperature 

30.5 

31.5.31 

30.5 

29 

28.5 

29.531 

31.5 

31 

30 

29.5 

29 

28  etc. 

2.  From  the  figure  in  Ex.  1  estimate  the  temperature  at 
at  11.30  A.M. ;  at  2.30  p.m.  ;  at  6.15  p.m. 

3.  At  what  time  of  clay  was  the  temperature  about  30.5°  ? 
29°  ?  Is  there  more  than  one  answer  in  each  case  ?  Find  all 
the  answers. 

4.  At  what  times  (about)  during  the  day  did  the  tempera- 
ture change  from  rising  to  falling  ?  from  falling  to  rising  ?  . 

5.  If  the  price  of  tea  is  35  cents  per  pound,  draw  a  figure 
to  show  the  cost  of  any  number  of  pounds. 

6.  From  the  figures  give  the  cost  of  3^  pounds;  of  41 
pounds;  of  —  l^-  pounds. 

[Hint.  —  Continue  the  line  below  and  to  the  left.] 

7.  How  much  tea  can  be  bought  for  84  cents?  for  $1.96? 
for  —  $1.05  ?     What  does  the  last  statement  mean  ? 

8.  The  population  of  the  U.S.,  as  determined  by  the  decen- 
nial census,  is  approximately  given  in  the  following  table : 


Year      .    . 

1790 

1800  1810 

1820  18.30 

1 

1840 
17 

1850 
23 

1860 
31 

1870 
39 

1880 
50 

1890 
63 

1900 
76 

Millions 

4 

5        7 

10  1  13 

Draw  a  graph  showing  the  population  at  various  times. 
9.  Estimate  the  population  in  1805;  in  1885;  in  1895. 
10.  When  was  the  population  about  15,000,000  ?  60,000,000? 
70,000,000  ? 


22 


MEASUREMENT    AND   EXPUES810X 


[Cii.  11 


11.  Represent  by  a  graph  the  simple  interest  at  6  %  for  one 
year  on  any  amount  of  money.  Choose  snms  at  intervals  of 
$  1000  from  -  $5000  to  +  $  5000. 

12.  What  interest  shall  I  have  to  pay  at  6  %  for  a  year's  nse 
of  S  2500?  $4500?  —$2500?  Interpret  the  last  statement 
in  the  light  of  Ex.  11. 

13.  What  principal  yields  as  interest  for  one  year  at  G% 
$195?    -$195?    -$105? 

14.  The  nnmber  of  inhabitants  of  the  United  States  of  school 
age  from  1870  to  1904  is  shown  in  the  Fig.  11.  What  was  the 
number  in  1870  ?  18S0?  1885?  1890? 


Nvtiiher 

of 

School  Age 


24000000 


S3  000  000 


22000000 
210000CO 
20000000 
19000000 
18000000 
17000000 
ICOOOOOO 
15000000 
14000000 
13000000 
12000000 


Yedn 


Fig.  11. 


15.   When  was  the  number  of  inhabitants  of  school  age  about 
12,000,000?  14,000,000?  10,000,000?  20,000,000? 


17-18] 


RELATIOxN  S  —  (i  R  A  PHS 


23 


16.  From  the  ti^nire  construct  a  complete  table  of  the  num- 
ber of  inhabitants  of  school  age  from  IbTO  to  1905  (every  year). 

17.  Draw  a  straight  line  graph  to  represent  the  correspond- 
ing readings  of  a  Centigrade  and  a  Fahrenheit  thermometer. 
(See  tables  at  back  of  book.) 

18.  Give  the  Centigrade  temperatures  corresponding  to 
Fahrenheit  temperatures  22°  ;  35° ;  -  17° ;  -  32°.  Give  the 
Fahrenheit  temperatures  corresponding  to  Centigrade  tempera- 
ture -25°;  -12°;  -5°;  20°;  85°. 

19.  What  temperature  has  the  same  reading  on  both  scales? 
The  teacher  should  also  have  each  student  take  some  problem  in  his 

own  experience:  the  number  of  inhabitants  in  his  town;  the  number 
of  pupils  in  his  school ;  the  earnings  in  his  father's  store;  the  price  of 
wheat  or  of  cattle;  the  amount  of  rainfall  in  various  years;  or  any 
other  familiar  instance  of  varying  quantity.  This  problem  should  be 
carefully  investigated  by  the  student,  and  the  figure  should  be  drawn. 
The  World  Almanac,  which  can  be  had  by  addressing  The  World, 
New  York  City,  contains  much  valuable  information  of  the  kind 
needed  in  such  problems. 

18.  Equation  and  Graph.  In  the  case  of  prices  of  a 
commodity,  such  as  in  the  example  in  §  17,  we  may  also 
represent  the  cost  of  any  number  of  pounds  by  an  equa- 
tion. Let  c  be  the  cost  in  cents,  p  the  number  of  pounds; 
then  if  butter  costs  30^  per  pound, 

6'=30j». 

Tills  equation  represents  the  same  thing  (cost  of  an 
amount  of  butter)  as  Fig,  10  ;  hence,  we  say  that  this 
equation  belotu/s  to  that  figure,  or,  briefly,  this  is  the  equation 
of  that  figure ;  and  that  figure  is  the  graph  of  this  equation. 

Notice  that  the  figure  may  easily  be  drawn  from  the 
equation.  Taking  various  simple  values  of  p,  we  get  the 
following  table  : 


Pounds     

I) 

1 

2 

3 

10 

etc. 

-1 

-2 

etc. 

Cost     -. 

0 

80 

60 

90 

.•300 

etc. 

-.30 

-60 

etc. 

24 


MEASUREMENT   AND   EXPRESSION  [Ch.  11 


If  we  plot  these 
points,  as  before,  we 
get  a  figure  lil^e  Fig. 
12. 

It  "will  be  evident  to 
the  student  from  the  fig- 
ure that  this  graph  is  a 
straight  line  through  the 
starting  point,  as  men- 
tioned in  §  17.  The  proof 
of  this  fact  is  given  in 
§  80,  p.  140. 

19.  Equations  of 
Prices.  We  proceed 
to  extend  the  sugges- 
tion of  §  IT  tliat  most 
price  curves  are 
straight  lines.  If  the 
cost  of  one  article  (or 
of  one  unit  of  meas- 
ure of  a  measurable 
commodity )  is  known, 
the  cost  of  an}^  num- 
ber is  given  by  the 
following  equation, 
where  c  is  the  total 
cost,  k  tlie  cost  of 
one,  and  n  the  num- 
ber bought: 

total  cost  =  (known    cost    of    one)  X  (number), 
or,  c=k-n. 

As  in  the  example  of  §  18,  the  graph  of  sucli  an  equa- 
tion is  always  a  straight  line  through  the  starting  point. 
(See  also  §  80.) 


7 

300  ^=$3                                           ' 

4 

1 

t 

1                   ^ 

6-                   7 

t 

1 

7 

L 

200  ci=rS2                             1 

7 

r 

1 

7 

r 

/ 

7 

r 

l\ 

100  ^=$1            / 

r 

/ 

7 

:x 

50  d     1  /[ 

^    7 

1 

l\ 

10(^7 

'  /                   lb 

/O                 5*                10* 

7 

r 

/ 

7 

r 

/ 

Fig.  12. 


18-2UJ 


RELATIONS  —  GKAPIIS 


25 


20.  Linear  Equations.  Straight  Lines.  The  student 
should  notice  that  ordinary  proportion*  might  be  used 
in  the  preceding  examples.  Other  examples  in  which 
proportion  might  be  used  lead  to  equations  similar  to  tliose 
above,  and  to  straight  line  graphs. 

Ex.  1.  If  a  man  walks  3  miles  per  hour,  and  if  d  denotes 
the  total  distance  (in  miles),  and  n  the  total  number  of  hours 
he  walks,  evidently  d  =  3  n. 

Ex.  2.  If  X  denotes  the  number  of  feet  in  a  certain  length, 
and  y  denotes  the  number  of  inches  in  the  same  length,  y=12  x. 

[Let  the  student  make  a  table  of  values  and  draw  a  graph  for  each  of 
these  examples.] 

Many  examples  arise  in  business  and  in  science  in  which 
ordinary  proportion  could  not  be  used  directly.  Such 
examples  may  lead  to  straight  line  graphs. 

Ex.  3.  If  the  cost  of  setting  the  type  for  a  circular  is 
$2.00  and  the  cost  of  paper  and  printing  is  |^  per  copy,  find 
the  cost  of  any  number  of  copies. 

Let  c  mean  the  total  cost  in  cents ;  let  n  be  the  number  of  copies. 
Then,  ^  =  In  +  200.  v 

In  order  to  plot  the  graph  of  the  equa- 
tion c  =  ^n  +  200,  let  us  call  Cj  the  cost 
of  printing  and  paper  alone,  then  r^  =  |  7i 
and  the  figure  is  a  straight  line,  as  above. 
Now  the  effect  of  the  cost  of  setting  the 
type  is  to  increase  the  price  by  200^,  no 
matter  what  the  number  of  copies. 
Hence,  the  figure  for  the  total  cost  is 
found  by  simply  raising  the  whole  figure 
by  an  amount  that  denotes  200  (»  on  the 
vertical  scale.  Hence,  the  final  figure  is 
also  a  straight  line. 


"  — r  —  ~  ""                 ----- 

1   - 

"■■   '^        I  ^  '       _  fxL,^' 

"  ■    K       1  ;    :  :  :      :avj^  ' 

+^__^,>^    ,:■=    ::^ 

-T---0   -       ini)  ■    -Jill) 

Fig.  13. 


*  It  is  assumed  that  the  student  is  familiar  with  ordinary  proportion  as 

treated  in  all  arithmetics.    If  not,  the  teacher  may  well  recall  it  to  his  mind. 


26 


MEASUREMENT   AND   EXrilESSlON 


[Cii. 


All  of  the  equations  of  §§  18,  19,  20  are  of  the  form 

y  =  Ax  +  / 

where  k  and  /  are  fixed  numbers.  The  graph  for  y  and  a;  is  a  straiglit 
line.  For  this  reason  any  equation  of  the  form  y  =  kx  +  I,  or  any 
equation  that  can  be  reduced  to  this  form,  is  called  a  linear  equation 
or  a  simple  equation. 


EXERCISES   III:  CHAPTER  II 

1.    If  11  tickets  to  an  entertainment  cost  $2.75,  how  much 
do  5  such  tickets  cost? 

Let  11  be  the  number  of  tickets  bought,  and  c  the  total  cost  (in 
cents)  ;  then  c  =  kn,  where  k  is  the  price  of  1  ticket  (in  cents).  On 
a  figure,  as  before,  plot  the  values  ?i  =  11,  c  =  275  (cents),  given  in 
the  problem,  at  P.  Join  P  and  O;  the  line  PC)  represents  the  cost 
of  any  number  of  tickets.  Go  5  units  to  the  right  of  0  and  then  up 
to  the  point  Q  ;  opposite  Q  is  the  point  which 
represents  the  number  125  on  the  vertical 
line.     Hence, 5  tickets  cost  125  cents. 

We  might  have  found  this  from  the  equa- 
tion. For  c  =  kn,  and  c  =  275  when  n  =  11 ; 
hence,      275  =  11  k, 


Y 


-V.           -t- 

M 

:          V 

i 

-F                      -J^i 

P 

^ 

:        '..  j"X    ' 

i     ip    ^ 

i       "^ 

5^  -^i^.-= 

100  (•=  SI-  /I  '!----- 1--~- 

;   '           /  '      '                  ' 

1  '  i         /■  '      1                  ' 

M          /:  1                      ' 

•  ^T    U)  i  T          '    ,--- 

or, 


k  =  25. 


X 


If  now  n  =  5,  we  have 

c  =  25  n  =  25  •  5  =  125, 

which  is  the  result  just  found. 

The  advantage  of  the  graph  is  that  it 
solves  all  sucli  problems  (approximately)  at 
once.  Again,  the  figure  quickly  shows  the 
answer  to  many  reverse  problems.  Thus,  if 
a  man  has  ^2.40,  how  many  tickets  can  he 
buy  ?  l^ook  for  the  point  for  210  on  the  verti- 
cal line,  pass  over  to  the  line  OP,  then  look  down  at  the  horizontal. 
It  is  clear  that  he  can  buy  only  9,  for  the  value  found  is  less  than  10. 

2.  Draw  a  graph  to  show  the  relation  between  kilograms 
and  avoirdupois  pounds.  AVhat  is  the  equation  expressing  the 
fornicr  in  terms  of  the  latter  ?     Compare  Ex.  2,  p.  25. 


Fig.  14. 


o,,j  UKLATIONS  — (iUArilS  27 

[Suggestion.  Let  x  denote  the  number  of  kilograms  in  a  given 
weight,  and  let  //  denote  the  number  of  pounds  in  the  same  weight. 
Then  //  =  2.2  x  x  (nearly),  suice  1  Kg.  =  2.2  lb.     See  Tables.] 

3.  Express  approximately  in  kilograms  3  lb.,  7  lb.,  11  lb. 
Express   approximately    iu    pounds   0.5   Kg.,  3  Kg.,  4.5  Kg. 

4.  What  is  the  equatiou  representing  the  area  A  of  a  rec- 
tangle whose  base  is  5  and  altitude  a'-      Draw  the  graph. 

5.  Find  A  in  Ex.  4  if  «  =  2  ;  if  ct  =  5 ;  if  a  =  3^.  Find  a  if 
^  =  15;  if  .1  =  271 

6.  Express  by  an  equation  the  relation  between  United 
States  dollars  and  English  pounds  sterling.  Compare  with 
Ex.  2.     Draw  the  graph, 

7.  Express  by  an  equation  the  relation  betw^een  the  differ- 
ence in  time  of  two  places  on  the  earth's  surface  and  the  differ- 
ence of  longitude.  Draw  the  graph.  Interpret  the  meaning 
of  positive  and  negative  values  of  each  difference. 

[Suggestion.  A  difference  of  15°  in  longitude  makes  a  difference 
of  1  hour  in  time.] 

8.  Expressing  longitude  east  of  Washington,  D.  C,  as 
positive,  and  longitude  Avest  as  negative,  find  the  time  corre- 
sponding to  noon  at  Washington  at  a  place  whose  longitude  is 
_l_30°5  —30°;  +25°;  —40°.  Find  the  longitude  of  a  place 
whose  time  differs  from  that  at  Washington  by  -\-  3  hr. ; 
-  3  hr. ;  -^  21  hr. ;  -  51  hr. 

9.  Eepresent  by  a  graph  the  distance  traversed  by  a  bicy- 
clist at  the  rate  of  8  miles  an  hour.  What  is  the  equation 
connecting  the  distance  and  the  time  ? 

10.  In  what  time  will  the  bicyclist  cover  2.3  miles?  3.2 
miles?     How  far  will  he  go  iu  li  hours?  in  3i  hours? 

11.  The  cost  of  setting  type  for  an  order  of  business  cards 
is  $0.75;  the  cost  of  cards  and  printing  is  i^  apiece.  What 
is  the  equation  for  the  cost  of  any  number  of  printed 
cards  ?  Draw  the  graph.  Estimate  the  price  of  150  cards  ; 
of  500  cards  ;  of  650  cards. 


28  MEASL'RKMENT   AND   EXPRESSTOX  [Ch.  1 1 

12.  Another  firm  offers  to  disregard  the  initial  cost  of  type- 
setting and  to  print  the  cards  for  \^  apiece.  What  is  the 
cost  of  any  number  of  copies  ?  DraAV  the  graph  on  tlie  same 
diagram  as  that  for  Ex.  9. 

13.  Which  firm  will  most  cheaply  print  100  cards  ?  500 
cards  ?  Which  firm  will  do  the  most  printing  for  $  1.25  ?  for 
$  2.00  ?  How  many  cards  will  be  printed  just  as  cheaply  by  one 
firm  as  by  the  other,  and  what  will  be  the  cost  of  the  printing  ? 

14.  The  initiation  fee  in  a  certain  club  is  $  3.00 ;  the  yearly 
dues  are  $2.00.  What  is  the  cost  of  membership  in  the  organ- 
ization for  any  number  of  years  ?     Draw  the  graph. 

15.  Find  the  cost  of  membership  in  Ex.  14  for  5  years ;  for 
12  years ;  for  15  years.  Find  the  length  of  membership  of  a 
member  who  has  paid  $11;  $29. 

16.  The  bicyclist  of  Ex.  9  has  a  distance  of  20  miles  to 
go.  Represent  by  an  equation  the  relation  between  the  time 
he  rides  and  the  distance  that  remains  for  him  to  travel. 
Draw  the  graph. 

17.  How  far  has  he  left  to  go  after  traveling  .82  honr  ? 
1.77  hours  ?  How  long  must  he  travel  in  order  to  be  within 
12  miles  of  his  destination  ?   4  miles  ? 

18.  Draw  in  the  graph  whatever  should  correspond  to  the 
bicyclist's  riding  past  his  destination.  How  long  after  start- 
ing will  he  have  left  —4  miles  to  travel  to  his  destination? 
-  12  miles  ? 

19.  Two  trains  are  running  at  the  same  rate  in  the  same 
direction.  If  x  is  the  distance  passed  over  by  one  train  in  any 
length  of  time,  and  y  the  distance  passed  over  by  the  other  in 
the  same  time,  what  is  the  relation  between  y  and  x?  Draw 
the  graph  showing  this  relation. 

20.  Two  trains  start  from  Chicago  and  run  at  the  same  rate 
in  oppofiite  directions.  If  x  and  y  represent  the  distances  of 
the  two  trains  from  Chicago  at  any  tinie,  express  the  relation 
between  y  and  x  by  an  equation  and  by  a  graph. 


20-21] 


Rf:LATIONS  — GRAPHS  29 


21.  Suppose  that  one  train  starts  from  Boston  when  another 
is  5  miles  ahead,  and  they  travel  at  the  same  rate  in  the  same 
direction.  AVhen  the  first  train  is  x  miles  from  Boston,  how 
far  is  the  other  from  Boston  ?  If  .'/  is  the  distance  of  the 
second  train  from  Boston,  wliat  is  the  relation  between  >/  and 
X  ?     Draw  the  graph. 

22.  A  problem  similar  to  the  preceding  leads  to  the  equa- 
tion ij  =  x  +  9.  Plot  the  graph.  When  x  =  3,  what  is  y? 
"When  y  =  17,  what  is  x  ? 

23.  From  a  time-table  of  some  railroad  between  two  im- 
portant cities,  construct  a  figure  to  show  the  movements  of 
trains  ;  mark  the  distances,  starting  from  one  city  along  the 
main  horizontal  line,  and  mark  the  times  of  day  from  12 
o'clock  of  one  day  to  12  o'clock  the  next  day  on  the  main 
vertical  line.  Then  plot  the  position  of  each  train  at  the 
time  shown  in  the  time-table. 

Plot  graphs  corresponding  to  the  following  equations  (plot 
only  positive  values  unless  you  know  how  to  perform  the 
necessary  operations  with  negative  values)  : 

24.  y  =  x  —  2.  When  y  =  7,  what  is  x  ?  When  y  =  5,  what 
is  x  ? 

25.  y  =  4:X—2.  Plot  only  points  for  which  x  is  positive. 
What  is  y  when  x  =  2L  ?  when  x=  31  ? 

26.  2  y  =  x  +  3.  (This  may  also  be  written,  y  =  ^x  +  ^.) 
What  is^/^vhen:«=l?  3?   5?    What  is  x  when  ?/=2  ?   3?  4? 

27.  5y  =  2 X  + 10.  What  is  y  when  a;  =  0?  2?  5?  What 
is  X  when  ?/  =  4  ?  6  ? 

21.  Other  Examples.  Some  examples  do  not  lead  to 
straight  line  graphs.  Those  which  give  a  linear  equation 
(§  20)  do,  of  course  ;  but  it  is  easy  to  make  examples  which 
lead   to   other  kinds  of  equations  and  to  other  kinds  of 

,   figures.     The  following  example  will  dispel  the  false  idea 

!  that  all  graphs  are  straight  lines. 


30 


MEASUREMENT   AND   EXPRESSION 


[Cii.  II 


Ex.  1.    What  is  the  area,  A,  of  a  square  in  terms  of  its  side, 
s  ?     Draw  the  graph. 

Solution.     If  A  is  the  area  (in  sq.  ft.),  and  .s  is  the  length  of  one 
side  (in  ft.),  we  have  A  =  .s-.    Giving  s  various  values,  we  find: 


Length  of  side  (s)  : 

0 

1 

2 

3 

4 

5 

etc. 

I 

3 
2 

etc. 

Area  of  square  (A)  : 

0 

1 

4 

9 

i 

J'oint  in  Fig.  15  : 

0 

A 

B 

C 

D 

E 

[Let  the  student  fill  in  the  blank  spaces  and  extend  the  table.] 


Y 

t 

».            et 

i 

1 

4 

20                    + 

t 

L 

bi 

1 

-+-       to           H 

1 

niT 

^  w 

^  fd 

% 

ii 

tt 

-^--<-m^ 

rI^ 

^ti 

tlX 

ATtl 

0  /       1     1        )     S  (s'fie) 

Scale: 

"  1  fl  h  rizontally 

1  kq.  ft.  vertically 

X 


Fig.  15. 


Plotting  these  values  of  s  and 
A,  we  obtain  a  graph  that  is  not 
a  straight  line  (Fig.  15).  This  fig- 
ure may  be  used  as  above ;  thus, 
if  the  side  is  2.3,  we  may  find  tiie 
area  by  going  out  2.3  on  the  hori- 
zontal line  to  the  right,  then  up  to 
the  curve  (R  in  Fig-  15),  then  over 
to  the  vertical  line ;  the  value  of  A 
is  seen  to  be  a  little  over  5 ;  really, 
A  =  5.29.  Again,  if  .4  =  11,  we 
can  find  s  =  3.3  (about)  ;  the  point 
is  7' in  Fig.  15. 

22.    Review    Exercises. 

The  following  exercises  lead 
to  equations  like  those  in 
§§  18,  19,  20,  that  is,  the 
graphs  are  straight  lines. 
When  one  of  the  quantities 
that  vary  is  given,  the  other 
can  be  found,  either  from  the 
equation  or  (approximately) 
from  the  figure.  The  two 
answers  should  be  the  same, 
except  for   the    sliglit   inac- 


curacy of  the  figure  ;  thus  the  figure  serves  as  a  check. 


21-22] 


REVIKW 


31 


REVIEW  EXERCISES    IV:  CHAPTER   II 

1.  Represent  by  a  graph  the  multiplication  table  for  mul- 
tiplier 7,  from  7  X  0  to  7  X  12.  Read  off  7  x  4.5 ;  7  x  2.6; 
7  X  3^. 

Determine  a  number,  n,  such  that  7  n  =  85.5  ;  7  n  =  3.5. 

2.  A  certain  grade  of  cloth  costs  20^  a  yard.  Represent 
in  a  figure  the  cost  of  any  number  of  yards.  Read  from  the 
figure  the  cost  of  3.5  yards;  —  o.o  yards;  1.4  yards.  (Write 
equations.) 

How  many  yards  cost  64  ^?    86^  ?   —  50^? 

3.  The  value  of  farm  property  in  the  United  States  is  given 
at  intervals  of  ten  years  from  1850  to  1900  in  the  following 
table : 


Year      .... 

18.-,() 

1860 

1870 

1880 

18!)0 

1900 

Billions  of  Dollars 

3.98 

7.98 

8.90 

12.18 

16.08 

20.44 

DraAv  the  graph;  estimate  the  values  for  intermediate  years, 
recording  your  results  in  the  form  of  a  table. 

4.  Represent  graphically  the  relation  between  the  circum- 
ference of  a  circle  and  its  radius.  Use  the  figure  to  determine 
a[i})roxiinately  the  radius  of  a  circle  whose  circumference  is 
3  ft. ;  22  cm. ;  35.5  units.     (See  Table.) 

5.  Letting  d  represent  one  degree,  what  is  the  circumfer- 
ence of  a  circle  in  terms  of  c?  ?  If  r  is  the  radius  of  the  circle, 
what  is  the  circumference  in  terms  of  r?  What  relation  there- 
fore holds  between  d  and  r  ?     Draw  the  picture. 

Express  as  a  multiple  of  the  radius  of  the  circle  an  arc  of 
57.3°;  100°;  172°.  How  many  degrees  are  there  in  an  arc 
twice  as  long  as  the  radius  ? 

6.  Repi-esent  \)y  a  picture  the  relation  between  pounds  and 
ounces.  Construct  and  solve  problems  in  the  reduction  of 
pounds  to  ounces  and  ounces  to  pounds.     Compare  Ex.  2,  p.  28. 

7.  Find  out  the  rate  per  $  100  valuation  of  your  city, 
county,  and  state  taxes.     Plot  a  figure  to  represent  the  tax  on 


32  MEASUREMENT   AND   EXPRP:SS1UN  [Cn.  II 

any  vahmtion  by  representing  the  valuation  along  the  horizon- 
tal line  ($100  to  one  large  space)  and  the  tax  along  the  vertical 
line  ($  1  to  one  large  space).  Draw  a  line  to  indicate  the  total 
tax.  Exj^ress  the  same  facts  by  equations.  Estimate  the 
taxes  on  a  house  and  lot  valued  at  $2750.  Find  the  value  of 
property  taxed  at  $  6.25. 

8.  A  man  rides  horseback  for  an  hour  and  a  half  at  the 
rate  of  8  miles  an  hour;  then,  leaving  his  horse,  he  walks  back 
at  the  rate  of  4  miles  an  hour.  Draw  a  picture  showing 
the  distance,  d,  of  the  man  from  his  starting  point  at  any  time,  t. 
How  long  does  the  trip  take  ? 

9.  Another  man  starts  at  the  same  time  from  the  same 
place  as  the  man  in  Ex.  8,  walking  at  the  rate  of  three  miles 
an  hour  in  the  same  direction.  How  long  after  the  start  will 
the  first  man,  returning,  meet  the  second,  and  how  far  from 
the  starting  point  will  the  meeting  take  place?  Solve  only 
graphically. 

10.  A  train  that  leaves  Kansas  City  at  10  a.m.  arrives  in 
St.  Louis  at  6  p.m.  The  distance  is  280  miles.  Find  the  aver- 
age speed.  If  t  denotes  the  time  (in  hours)  after  leaving  Kansas 
City,  and  d  denotes  the  distance  (in  miles)  from  Kansas  City, 
express  by  an  equation  and  by  a  graph  the  relation  between 
d  and  t.  Find  d  when  t  =  2i.  Find  t  when  d  =  105 ;  when 
d  =  200. 

11.  A  train  leaves  New  York  at  2  p.m.  and  arrives  in  Chicago 
(940  mi.)  at  5  p.m.  the  next  day.  Express  by  an  equation  and 
by  a  graph  the  relation  between  distance  and  time.  When 
should  the  train  reach  Buffalo  (430  mi.  from  New  York)  if  its 
speed  is  never  changed  ? 

12.  Plot  the  graphs  of  the  equations  y  =  2x  —  9,  y  =  x  —  3, 
on  the  same  diagram. 

What  pair  of  values  of  x  and  y  satisfies  both  equations,  i.e. 
belongs  to  both  figures? 

13.  Proceed  as  in  Ex.  12  in  case  of  the  equations: 

y  =  10  —X,   y  =  x  —7. 


?2]  SUM.MARY  33 


SUMMARY   OF    CHAPTER  II:    MEASUREMENT    AND   AIDS   IN 
EXPRESSION,   pp.  13-32 

Part  I.   Simple  Measurkmknt;  Negative  Quantities,  pp.  13-16. 
Simple  Scale  of  Pusitice  Xumhers:    measurement  by  rulers. 

§  1-2,  p.  13. 
Introduction  of  Negative  Numbera  :  thermometers  ;  readings  below 

zero;  other  opposites.  §§  13-14,  pp.  13-14. 

Representation    of   Negutires :    vertical   scale    like   thermometers ; 

horizontal  scale,  positive  to  tlie  right. 
Addition  and  Subtraction  (fa  Positive  Number  :  motion  up  or  down 

on  vertical  scale,  forward   or  backward  on  horizontal  scale. 

Exercises  I.  §  15,  pp.  15-16. 

Part  II.   Relation  Between  Two  Quantities  ;  Graphs,  pp.  17-32. 

Temperature  Curve:  development  by  plotting  points,  estimation  of 
temperature  at  given  time;  time  for  given  temperature  ;  graphs. 

§  16,  pp.  17-20. 

Price  Curves:  usually  straight  lines.    Exercises  II,  §  17,  pp.  20-23. 

Graph  of  an  E(/uatiori :  each  pair  of  numbers  from  the  curve  satis- 
fies the  equation.  §  18,  p.  23. 

E(iuations  (f  Prices :  c=  kn;  straight  line  graph  through  starting 
point.  §  19,  p.  24. 

Linear  Equation  :  ordinary  proportion,  y  =  kx  -\-  I ;  straight  line 
graph.     Exercises  III.  §  20,  pp.  25-29. 

Other  examples:  equations  leading  to  figures  not  straight  lines; 
graph  for  area  of  square.  §  21,  pp.  29-30. 

Review  Exercises  for  Chapter  II:  relations  between  quantities. 
Exercises  IV.  §  22,  pp.  30-32. 


CHAPTER   III.     ADDITION   AND   SUBTRAC- 
TION;   SIMPLE   EQUATIONS 

PART   I.     GENERAL   RULES   FOR   OPERATION  ; 
PARENTHESES 

23.  Extension  of  the  Operations.  The  student  lias  seen 
r§  15,  p.  15)  how  to  add  and  subtract  in  several  new  cases. 

-4^  —3-2-1  0  1  2  3  4  5 

-j— I — I — I — I — I — I — 1 — I — I — i— t — I — I — I — I — I — I — I — I — I — I — I — I — I — >— I — < — I — I — I — \ — I — I — i — I — l- 

Fi<;.   Ki. 

We  have  seen  that  we  go  forward  to  add  a  positive 
number,  baehvard  to  subtract  a  positive  number  ;  thus, 
2  +  3  =  5,  —1  +  3  =  2,  2  —  5  =  —  3,  etc.  Let  us  carry 
out  these  processes  systematically,  proceeding  to  perform 
addition,  subtraction,  etc.,  in  cases  not  mentioned  in 
elementary  arithmetic.  In  doing  so  we  shall  think  of , 
addition,  for  example^  as  the  same  operation  as  that  used 
in  arithmetic^  exferided  notv  to  all  the  numbers  tve  know,  the 
extension  being  carefully  made  in  order  that  the  most  essen- 
tial properties  of  el&i.ientary  addition  remain  true ;  these 
properties  are  mentioned  belotv. 

24.  Properties.  One  important  characteristic  of  addi- 
tion is  the  fact  that  the  numbers  added  may  be  taken 
in  any  order  without  changing  the  result. 

Thus, 3  +  4  =  4  +  3,  for  both  equal  7.  The  same  thing-  is  true  in 
all  additions  of  elementary  arithmetic. 

Instead  of  writing  down  every  possible  example,  we  may 
say  that  a  +  6  =  6  +  «,  if  a  and  b  mean  any  two  numbers 
whatever.  •     ■• 

34 


GENERAL   RULES  36 

Tliis  fact  is  often  used  to  check  the  correctness  of  additions,  for  if 
a  problem  is  solved  by  adding  the  numbers  in  two  different  orders,  the 
results  shoidd  be  the  same.  Thus,  one  often  adds  a  column  from  the 
bottom  upward,  and  then  also  from  the  top  downward. 

The  most  fuiidainental  properties,  including,  for  con- 
venience, the  properties  of  multiplication,  are  the  fol- 
lowing : 

I.    a  +  b  =  b  +  a. 

This  is  mentioned  just  above.     Example  :  3  +  4  =  4  +  3. 

II.    a  X  b  =  b  X  a. 

This  has  to  do  with  multiplication.     Example  :  5x7  =  7x5. 

III.  a+{b +  c)  =  (a  +  b)  +  c.  . 

Example  :  2  +  (7  +  4)  =  (2  +  7)  +  4,  that  is,  2  +  (1 1)  =  (9)  +  4. 

This  rule,  also,  is  often  used  in  adding  columns  of  figures;  thus, 
instead  of  adding  the  numbers  one  by  one,  we  may  add  groups,  if 
convenient;  for  example,  if  a  figure  7  follows  a  figure  3  we  add  10, 
instead  of  adding  7  and  then  3. 

IV.  a  x(6  X  c)  =  (a  X  6)x  c 

Example  :  3  x  (5  x  4)  =  (3  x  5)  x  4,  that  is,  3  x  20  =  15  x  4. 
This  is  often  used;  thus,  (5  x  8)  x  6  is  easier  than  5  x  (8  x  6). 

V.    a(b  +  C)  =  ab  +  ac. 

Example  :  5  (3  +  7)  =5  x  3  +  5  x  7,  that  is,  5(10)  =  15  +  35. 
This  is  often  used ;  thus,  52  x  7  =  50  •  7  +  2  •  7,  which  is  easier. 
This  is  really  the  principle  of  multiplication  taught  in  arithmetic. 

These  rules  are  called  axioms,  because  tliey  are  merely 
assumed  to  be  true  ;  there  is  no  pretense  of  proving  them. 
We  take  these  rules  in  preference  to  any  others  because 
experience  has  shown  that  they  are  the  simplest  and  most 
useful.     See  also  the  footnote  on  p.  56. 

These  rules  are  frequently  named  as  follows  :  T.  Commutative  Law 
of  Afldition  :  TI.  Commutative  Law  of  Mulliplicntion  :  III.  Associative 
Law  of  Aililition  :  IV.  Associative  Law  of  Multiplication;  V.  Dis- 
trilintire  Luiv.     We  shall  not  often   use  these  names. 


36  ADDITION    AND   SUBTRACTION  [Ch.  Ill 

EXERCISES    I  :    CHAPTER    III 

Show  that  the  above 
rules  hold  if :  Ky  means  of  Rule  V : 

1.  a  =  7,  ^  =  3,  c  =  5.  5.  Multiply  28  by  5. 

2.  a  =  0,  b  =  2,  c  =  7.  6.  Multiply  63  by  8. 

3.  a  =  f,  b  =  l,  c=l  7.  Multiply  75  by  i 

4.  a  =  l,  b  =  2,  c  =  3.  8.  Multiply  6|  by  7. 

Use  the  above  rules  in  order  to  perform  the  operations 
indicated : 

9.    3  X  (4x31). 
Solution.     3  x  (4  x  31)  =  3  x  (3i  x  4)  =  (3  x  ^)  x  4  =  10  x  4  =  40. 

10.  121%  of  $200. 
Solution. 

12^%  of  1 200  =  (121  X  jij)  X  f  200=  12^  X  (^i,  x  i$200)  =  i$25. 

11.  2 +  (8 +  7).  14.    [8+(12  +  4)]xo. 

12.  5  X  7  +  5  X  3.  15.   41  x  41. 

13.  16|%  of  $300.  16.   53  X  53. 

25.  Addition  and  Subtraction:  Negatives.  As  mentioned 
in  §  23 : 

(1)  To  add  a  positive  number  we  go  forwards  on  the  scale. 

(2)  To  subtract  a  positive  number  we  go  backwards  on  the 
scale. 

We  now  add  to  these  the  following : 

(3)  To  add  a  negative  number  we  go  backwards  on  the 
scale,  by  the  amount  indicated  by  the  number. 

Thii8,  n  dollars +  3  dollars  debt  =  9  dollars +  (- 3  dollars)  =  6 
dollars,  which  is  the  same  as  9  dollars  —  3  dollars.  Likewise, 
9  (I  +  (—  3  dj  =  6  d  —  Q  d  —  3  d,  no  matter  what  d  ineans.  This 
scheme  is  useful.  The  student  will  see  also  that  in  this  process  the 
rules  mentioned  above  hold  pjood.  For  example,  {9  d)+  (—  '■]  d)  = 
(  — 3  f/)  +  (9  d)  by  I  ;   this  is  true,  for  eacii  sum  is  (3  d. 


04_ou]  (iENEKAL    RULES  37 

H — I — (— I — f— t— < — I — I — I- 


H — e 


-^  -2  -^  0  1  2    .  f    ,    ,^    f   .    ,    ,f 


To  add  a  positive  number  )  /-,     j?         „   ^„ 

—  4—3-2-1  0  1  2  3  4  5 

I    I    I     I     I    I    I    I    I    I    .    I    I    I    I    I    I    I     I     I     I    I    I     I    I     I    I    I    I     I    I    '    I    I    '-I    1' 


To  subtract  a  positive  number  )    ^     ,       ,  , 

or  ^     ^^  f.  .  }.  G^(j  backwards, 

to  add  a  negative  n^tmber  j 

Fig.  17. 

(4)  To  subtract  a  negative  number  we  go  forwards  on  the 
scale,  by  the  amount  indicated  by  the  number. 

Thus,  subtracting  (or  removing)  a  debt  amounts  to  increasing  the 
wealth  of  the  man  who  owed  it.  If  you  have  ten  dollars  and  owe 
four  dollars,  your  total  wealth  is  six  dollars.  But  if  some  one  pays  the 
debt  for  you,  you  will  really  have  ten  dollars ;  i.e.  Qd-  {-^  d)  - 10  d. 

(5)  To  subtract  any  number.,  change  its  sign  and  then 
add  the  resulting  number.  This  rule  is  very  important,  as 
will  be  seen  later. 

26.    Addition  and  Subtraction  of  Several  Numbers.     In 

business  and  ordinary  life  we  often  need  to  add  several 
numbers  some  of  which  are  positive  and  some  negative. 
Thus,  if  a  merchant  has  several  debts  and  also  several 
different  amounts  of  money  and  valuable  goods,  all  the 
debts  are  added  together  to  find  the  total  debts,  and  the 
values  of  all  the  money  and  goods  are  added  together  to 
find  the  total  "assets."  The  difference  between  these 
amounts  represents  the  real  wealth  :  it  is  a  debt  {i.e. 
negative  wealth)  if  the  total  debts  are  greater  than  the 
total  assets;  it  is  real  wealth  {i.e.  positive),  if  the  total 
assets  are  the  greater.  Some  examples  are  given  below 
in  which  this  process  proves  useful. 

When  several  numbers  are  to  be  added  together.,  ive  add  all 
the  positive  numbers  to  make  a  positive  total.,  then   all  the 


:i8  ADDITION   AND   SUBTRACTION  TCh.  Ill 

i 

negative  numbers  to  make  a  negative  total;  the  sum  sought 
is  the  difference  of  the  amounts  of  these  totals  icith  the 
sign  of  the  greater  one  prefixed  to  it. 

Thus,  the  sum  of  Q  d,  —  7  d,  4  (/,  —  10  d,  25  d,  is  found  as  follows  : 

Q  d  +  4  d  +  25  d  =  35  (/,  the  total  of  positive  numbers. 

—  7  ^/  —  10  (/  =  —  17  d,  the  total  of  negative  numbers. 

35  d  +  (—  17  <l)  =  35  (/  —  17  d  =  18  d,  the  final  answer. 
Again, 

3  ^/  +  (  _  5  ^/)  +  8  (/  +  (  -  12  (/),  is  found  as  follows  : 

3rf  +8  d  =  lid,  total  + 

-  5  d  -12d  =  -  17  d,  total  - 

11  ,/  4-  (_  \7  d)  =-(17  -  11)^/  =  -  6r/,  the  final  answer. 

EXERCISES   II  :  CHAPTER    III 

Perform  the  operations   indicated.     In   each   case   draw   a 
li'nire  showing  the  scale. 


12  d). 

-5  b). 


1. 

10-(-8). 

9. 

42  _  (-  36). 

2. 

3  -  4. 

10. 

$  215  -  f  472. 

3. 

4 +(-3). 

11. 

-  65  d  +  39  d  -  ( 

4. 

2 -(-2). 

12. 

17  a- -(-35  a-). 

5. 

-3-(-4). 

13. 

rjb  +  (-7b)  +  ( 

6. 

-3  +  (-4). 

14. 

i}a  -  (-Oa). 

7. 

8-(-7)  +  (- 

-5). 

15. 

{},,  —  {Ga). 

8. 

7-0. 

16. 

Oa-  (-6a). 

17. 

What  will  be 

the 

total 

wealth  in  money  of  a  i 

has  $  1500  in  the  bank  and  owes  %  1200,  if  by  a  transfer  of 
l)roperty  he  cancels  $  750  of  a  debt  ? 

18.  A  weight  of  30  lb.  is  attached  to  three  small  balloons 
which  pull  up  respectively  10  lb.,  15  lb.,  and  20  lb.  What  is 
the  upward  pull  of  the  whole  device  ?  Hence,  what  may  we 
say  is  the  weight  (downward  force)  of  the  whole  device? 
Express  the  problem  in  terms  of  positive  and  negative  numbers, 
using  downward  force  as  positive. 


20-27]  GENERAL    RULES  39 

19.  A  vessel  capable  of  making  10  miles  an  hour  is  opposed 
by  a  current  of  0  miles  an  hour.  By  hoisting  sails,  use  is 
made  of  the  wind  blowing  in  the  same  direction  at  8  miles  an 
hour.  What  is  the  speed  at  which  the  vessel  moves  ?  Express 
in  terms  of  positive  and  negative  numbers. 

20.  A  vessel  is  making  use  of  both  sails  and  steam;  the  latter 
alone  will  propel  it  at  15  miles  an  hour,  the  wind  opposes  the 
motion  at  the  rate  of  10  miles  an  hour,  and  an  opposing 
current  is  flowing  at  2  miles  an  hour.  What  is  the  speed  of 
the  vessel  ?  The  sails  are  lowered.  What  now  is  the  speed  ? 
Express  in  terms  of  positive  and  negative  numbers. 

,  27.  Addition  and  Subtraction  of  Negative  Numbers.  By 
itlie  rules  above  : 

(1)  Adding  a  negative  number  is  equivalent  to  snhtraet- 
'ing  a  'positive  number  of  equal  amount;  they  both  mean 
going  backward  on  the  scale  by  the  same  amount. 
,  (2)  Subtracting  a  negative  number  is  equivalent  to  add- 
ing a  positive  number  of  equal  amount;  they  both  mean 
going  forward  on  the  scale. 

Thus,  5+ (-2)  =  5  -  2  =  3, 

I  5_(_o)  =  5  +  2  =  7, 

land  so  on. 

I  This  holds  eqiialh^  well  for  dollars  (d  is  used  below)  or  for  any 
iother  unit  quantity : 

I  5d  +  (--2  d)  =  5  d  ~2d  =  ;}  d. 

I  5  d  -  (-  '2  d)  ^  5  d  +  2  d  =  7  d. 

I 

i  It  follows  that  additions  and  subtractions  of  nega- 
tive numbers  may  be  turned  into  subtractions  and  addi- 
itious  of  positive  numbers  by  changing  the  sign  of  the 
Igiven  quantity. 

ffence,  we  never  really  distinguish  between  such  expres- 
\sions  as 

[     5  +  (—  2)  and  5  —  2  ;   or,  in  general,  a-{-(—  b^=  a—  b; 


40 


ADDITION    AND   SUBTRACTION 


[Ch.  hi 


nor  between 

5  —  (  —  2)  and  5  +  2  ;  or,  in  general^  a  ~(^—b')  =  a  +  b. 

We  also  remark  that 

5  —  (  +  2)  means  5  —  2  ;  or,  in  general^  a  —  (^-\-  b)=a  —  b. 

With  this  understanding,  subtractions  or  additions  of 
either  positive  numbers  or  negative  numbers  may  be 
performed  as  above,  by  first  thinking  of  all  subtractions 
as  turned  into  additions. 


EXERCISES    III  :  CHAPTER   III 

1.  A  man  has  S215  in  one  bank,  $134  in  another,  and 
goods  worth  S1U50;  there  stand  against  him  a  debt  of  $75, 
and  a  mortgage  of  $750.  Using  d  for  one  dollar,  find  his 
total  wealth. 

2.  Letting  d  stand  for  one  dollar,  complete  the  following 
table,  which  represents  increases  (  +  )  and  diminutions  (  — )  of 
the  wealth  of  a  family  of  three  brothers : 


Jan. 

Feb. 

U\R. 

ToT.\L    FOR    TllREK    MoNTIIS 

A.     .     . 

912  d 

-Ud 

36  d 

B.     .     . 

—  721  d 

255  d 

-  1000  d 

C.     .     . 

2200  rf 

-  133  rf 

756  rf 

Total  . 

3.  A  weight  of  20  g.  (grams)  in  a  certain  piece  of  mechanism 
is  to  be  raised  by  two  wires  joulling  upward  by  6  g.  and  8  g. 
respectively,  and  a  spring  exerting  an  upward  pull  of  10  g. 
With  what  upward  force  will  the  weight  rise  if  the  smaller 
wire  has  been  broken?  Will  the  total  force  be  exerted  up  or 
down,  and  what  will  be  its  magnitude  ? 

4.  If  p  denotes  one  pound,  the  weight  of  the  basket  of  a 
balloon  is  320  p,  of  the  nistruments  68  p,  of  each  of  9  sand 


27]  GENERAL    RULES  41 

bags   100  p;    the    weight   of    the    balloon    itself  is  —loOOjx 
What  is  the  total  weight  of  balloon  and  contents  ? 

5.  Let  m  denote  miles  an  hour.  A  boat  using  both  sails 
and  steam  is  urged  forward  by  its  steam  power  at  the  rate  of 
12  m,  and  by  the  wind  at  the  rate  of  l  m.  It  sails  against  a 
curi-ent  flowing  at  the  rate  of  5  vi.  What  is  its  rate  of 
progress  ? 

Perform  the  following  additions : 

6.  12  +  9  +  (-7)  +  (-13)  +  (-2). 

7.  (-7)  +  (-9)  +  r,  +  (-2)  +  20. 

8.  8d  +  7  d  +  {-2()d) +6d+(-3d). 

9.  (_  7  x)  +  4  .r  +  10  .r  +  (-  5  x)  +  12  x. 


10.  (a)  15 

(b) 

—  5 

(c) 

Ih 

(d)        2  z 

-18 

-4 

-46 

-Sz 

32 

-3 

86 

5z 

-17 

2 

12  6 

9z 

-    5 

(/) 

10 
4.3 

(9) 

-36 
16  i> 

-17  z 

(e)          6 

(h)   -12/ 

-7 

-7.25 

-18p 

-10/ 

-9 

-6.8 

12p 

75/ 

-15 

10.53 

Sp 

-68/ 

20 

-8.2 

-15p 

-      / 

;      11.    The  averac/e  of  two  numbers  is  their  sum  divided  by  2. 
i  Find  the  average  of  4  and  6 ;  of  —  4  and  6 ;  of  10  and  15 ;  of 

- 10  and  15;  of  -  4  and  -  6 ;  of  - 10  and  -  2 ;  of  5  d  and  7  d ; 

of  7 d  and  —5d;  of  —7c?  and  —  5 d. 

12.  The  average  of  several  numbers  is  their  sum  divided  by 
the  number  of  numbers  added.  Find  the  average  of  2, 3,  4,  and 
5 ;  of  -  2  ,  3,  -  4,  5 ;  of  5  d,  -  7  c?,  and  3  d ;  of  -lOd,  -  12  d, 
and  —4:d. 

13.  In  Ex.  2,  find  the  average  gain  per  month  of  A  during 
the  three  months  given ;  of  B ;  of  C ;  of  the  whole  family. 


42 


ADDITION   AND   SUBTRACTION 


[Cii.  Ill 


14.  In  Ex.  2,  find  the  average  gain  of  the  three  men  during 
January ;  during  February  ;  during  March.  Find  the  average 
gain  for  the  three  months. 

15.  Find  the  average  of  the  numbers  added  in  Ex.  6;  in  Ex.  7; 
Ex.  8 ;  Ex.  9 ;  in  each  part  of  Ex.  10. 

16.  In  Ex.  2,  how  much  above  the  average  earnings  during 
January  were  A's  earnings  during  that  month  ?  How  much 
were  C's  above  the  average?  How  much  were  B's  below  the 
average  ?  Can  you  state  that  B's  earnings  were  a  certain 
amount  above  the  average  by  using  a  negative  number  ? 

17.  A  business  house  gains  $5000  one  year,  $2000  the 
next;  it  loses  $1500  the  third  year,  and  $3000  the  fourth; 
the  fifth  year  it  gains  $  1000.  What  is  tlie  total  gain  for  the 
five  years?     Find  the  average  yearly  gain  for  the  five  years. 

18.  In  Ex.  17,  how  much  above  the  average  gain  per  year 
was  the  gain  during  the  first  year  ?  during  the  second  ?  third? 
fourth?  fifth? 

19.  A  business  house  during  three  successive  years  loses 
$500,  $300,  and  $350.  What  is  the  total  loss?  the  average 
loss  ?  What  is  the  total  gain,  if  the  results  are  expressed  as 
gain?  the  average  gain?  State  the  result  as  a  problem  in 
averaging  negative  numbers. 

20.  The  results  of  a  business  for  five  years  are  as  follows : 


First  Yeah 

Second  Year 

TiiiKi)  Year 

Fdl  TRTII    Ykak 

FiKTii   Vi:aii 

Gain 

1 1732 

$  3000 

Loss 

f  5251 

f  2572 

1 135 

What  is  the  total  net  f/am  ?    the  average  gahi?     State  the 
result  as  a  jjroblem  in  averaging  positive  and  negative  numbers. 

21.    Find  the  average  of  the;  lollowing  tcinpcratau'es: 
-t  no  <>  n°  1  o  '>o  •>  ro 


27] 


GENERAL   RULES 


43 


22.  Wliat  is  tho  average  temperature  in  the  example  of  §  16, 
[I.  17  ■/    ill  tlie  example  on  p.  18  ? 

23.  A  rifle  gives  the  ball  the  muzzle  velocity  of  1200  feet 
per  second.  Fired  directly  backwards  from  a  moving  car,  the 
actual  velocity  of  the  ball  is  lloG  feet  per  second;  how  fast  is 
the  car  moving  ? 

24.  Fill  in  the  spaces  left  blank  in  the  following  account  of 
the  gains  of  a  merchant  at  his  three  stores  : 


First  Store 

Second  Store 

Third  Stoke 

Total 

1902 
11)03 
1901 
1905 
1900 

$5,000 

$;3,ooo 

-  $       50 

-  $2,250 

$  10,500 

$    8,000 

$    6,250 

-  $    1,050 

IS 

$10,000 
$11,525 

$ 
-  $  4,500 
$    1,000 

Total  gain  of 
live  years 

$8,700 

$ 

$28,225 

Average  gain 
per  year 

$4,540 

25.  Tlie  total  weight  of  a  balloon  and  basket  containing 
a  man,  apparatus,  sand  bags,  etc.,  and  small  balloons,  is 
—  2o,000p,  where  "p"  denotes  one  pound.  The  man  weighs 
150  p,  the  basket  and  apparatus  50/),  the  ten  sand  bags  eaeli 
weigh  75  p:  and  the  ten  balloons  each  —5^.  What  is  the 
weisrht  of  the  large  balloon  ? 


26.    Carry  out  the  following  subtractions : 
(a)       15         (c)  -25         (e)       17         (g)       17 


(&) 


-30 


-17 


(0 


ox 
—   Ox 


-25 

(d)       25 

(/) 

-17 

('0 

-17 

O) 

-15p 

17 

-17 

-25 

25 

-25p 

44  ADDITION   AND   SUBTRACTION  [Ch.  Ill 

27.  Reduce  the  following  expressions  to  simpler  forms : 
(a)   7x-\-{-llx)-{-12x)  +  (-5x)-6x. 

(5)    _25&-(-306)  +  126-166-(-6). 

(c)  10d  +  7d-(-5d)  +  (-30d). 

(d)  (_5)  +  (-5)  +  (-5)  +  (-5)  +  (-5). 

(e)  i-3k)-\-(-Sk)  +  (-3k)  +  (-3k). 
(/)  125z-(-375z)-{-(-752z)-502z. 

(g)    13  r  -  50  r  -  (-  65  7-)  +  73  r  -  11  r  -  (-  12  r). 

[Check  each  exercise  by  substituting  some  number  for  the  letter 
that  occurs  both  in  the  exercise  as  given  and  in  your  answer.] 

28.  Addition  of  Monomials.  We  have  done  one  thing 
above  which  should  be  stated  carefully.  Having  given 
9d+^d,  we  write  as  answer  12  d ;  and  this  is  correct 
whether  d  means  dollars,  dimes,  or  anything  else.     Thus, 

9d+Sd  =  (9  +  3)d  =  V2d, 

Ux+-nx=[U-i-i-n)]x=^x, 
and  so  on. 

Rule.  To  add  terms  that  have  a  common  factor,  add, 
the  coefficients  of  that  factor  ;  the  mm  is  the  sum  of  the  co- 
efficients times  the  common  factor. 

This  is,  in  fact,  merely  another  statement  of  V,  p.  35  ;  e.g. 
9(/  +  3r/  =  (9  +  3)^/=  1-2  d. 

The  same  rule  applies  for  the  sum  of  any  number  of 
terms. 

Thus,  (6  d)  +  (-1  d)  +  (4d)+(-10  (Z)  +  (25  d)  = 
[(j+(_7)  +  4_|-(_10)+2r)]  t?  =  18(/,  as  on  pp.  37-38. 
(^r  again,  1  x-{-(-12x)  +  (j x  +  4x  =  ll  +  (-ny  +  ij  +  ^'jx 
=  fix  ;   etc. 

This  rule  applies  no  matter  how  complicated  the  units 
may  be;  thus,  if  we  are  talking  of  special  objects  it 
remains  true  that  six  of  them  plus  three  of  them  =  nine 


27-29]  GENERAL   RULES  45 

of  them.  Or  again,  if  we  are  speaking  of  expressions, 
however  complicated,  such  as  b"^!-^,  it  remains  true  that 
6  b\v^  +  3  62^«  =  9  62^3,  etc. 

We  use  this  rule  for  the  addition  of  siynilar  terms. 
Let  the  student  read  again  and  state  the  definition  of 
similar  terms,  §  8.  By  proper  choice  of  a  common  factor 
many  terms  not  similar  as  a  whole  may  be  made  similar  in 
part  of  the  letters.  Thus,  by  V,  p.  35,  ax  +  hx  =  (^a-\-  b)x  ; 
aba^i/  +  cdx'^y={ab  +  cd^cc^y  ;  etc.  The  coefficient  is  chosen 
such  as  to  make  the  terms  similar. 

A  simple  check  on  the  answer  is  obtained  by  pntting  in  various 
numbers  at  random  in  the  place  of  the  letters.  Thus,  if  ft  =  3  and 
X  =  2  in  the  preceding  example,  then  6^  _  g^  a;3  _  g,  b^x^  =  72  ; 
and  6  h'^x^  =  432,  3  bV  =  216,  9  h^x^  =  648.  Now  432  +  216  =  648 ; 
if,  instead,  we  had  found  an  untrue  addition  here,  we  should  know  we 
had  made  an  eiTor  in  the  work  above. 

Such  a  check  is  not  an  absolute  proof  of  correctness,  however;  thus, 
2a;  +  3a;  =  4jc+lis  not  correct  in  general,  although  we  find  it  is  cor- 
rect for  one  choice  of  x,  namely,  for  x  =  1. 

29.  Subtraction  of  Monomials.  Since  subtracting  a 
quantit}'  is  equivalent  to  adding  its  negative,  we  may 
subtract  by  turning  all  subtractions  into  additions  of  the 
opposite  kind  of  quantity. 

Thus,  6  dollars  +  (-  3  dollars)  =  6  dollars  -  (+  3  dollars),  and 
6  dollars  +  (+  3  dollars)  =  6  dollars  —  (-  3  dollars),  as  has  been 
noticed  above  ;  or,  in  general, 

fa+  (-h)  =  n-h, 
\a+  (+h)=a-  (-b). 

The  rule  given  above  holds  for  subtraction  also,  it  being 
understood  that  to  subtract  a  quantity  means  the  same  as  to 
add  its  negative.     (See  §  25,  p.  37.) 


46                       ADDITION    AND   .SUBTRACTION  [Cii.  Ill 

EXERCISES   IV:  CHAPTER   III 

Perform  the  following  additions  : 

1.      6d                 2.    -9  A;            3.        4:k^x  4.        3abz 

—  Id                      —'3k                      7 k^x  — 10  abz 

18  d                         13  k                -13Fx  12  abz 

—5d                         —k                      3k^x  —15  abz 


5.    -3.5xhf  6.         2ir/«2  ^         ^^^^^ 

-1.1  xnf  -Hgf"  '  T 

2.9  a^y  2§ge  -2.7^ 


8.    -lA2ab^c  9.        O.W7n%n  +  r)      10.  as-t^ 

2.14  ab'c  -  3.24  m%n  +  r)  -  bsH^ 

- 1.98  ab^,  -21    m\n  +  r)  csH^ 

11.     6(ax''  +  bi/)  12.   -3C/:c4-r/.'/+/'2)     13.  4(-o2)  +  &r/) 

-  8  {ax'  +  &?/2)  -  8  (fx  +  gjj  +hz)          -  5  (  -  aji  +  /x/) 

3(ax'  +  by^)  7  (fx  +  gy  +  hz)  -  ( -  ap  +  /j?) 

Perform  the  following  subtractions  : 

14.    12bcyz      15.  -7.^7/^"      16.        3.81  /-^  17.    lm{m—l) 

- 10  bcyz  3x"yz-             -4.21r'''  ^m(:m-l) 


18.     -5-3-  19.    -l.T-^  20. 

R'  b- 


K'- 

-i.) 

12  ri- 

-i) 

21.   A{x  +  y  —  2)         22.       6yy/m^  +  y'''       23. 


.  3  (a;  +  y  _  2)  -  4  ?/  V w< '  + .'/' 


-  1.3  VI  - 

-f 

-l.oVl- 

-.'/ 

29-30]  GENERAL   KILKS  — PARP:NTHESES  47 

24.  Subtract  the  last  quantity  in  each  of  the  first  13  exercises 
from  the  sum  of  all  that  precede  it  in  the  same  exercise. 

25.  Average  the  quantities,  5  abx,  —  7  abx,  10  abx,  —  12  abx. 

26.  Average  the  quantities  in  each  of  the  first  13  exercises. 

27.  Average  the  quantities,  7  (6-x- +  «-/),  —  10  (b'-x^ -\- a'-y-), 
b-x-+d-U'-,  and  6{b-x^+a^y^). 

28.  Subtract  the  first  quantity  in  each  of  the  first  13  exercises 
from  the  average  of  the  rest  of  the  quantities  in  the  same  ex- 
ercise. 

30.    Grouping  in  Addition.     If  longer  expressions  are  to 
be  added  together,  we  use  the  fact  that  the  order  in  which 
terms  are  added  is  immaterial,  stated  in  Axioms  I  and  III, 
p.  35: 
I  a  +  b  =  b  +  a; 

III  a  +  (5  +  6')  =  (rt  +  b}  +  o. 

This,  ive  may  inclose  any  number  of  terms  in  parentheses 
preceded  by  the  sign  +  (plus);  likeivise  we  may  remove 
parentheses  preceded  by  the  sign  +  •  This  means  we  may 
add  the  terms  inside  a  pair  of  parentheses  together  before 
we  add  their  sum  to  the  others,  or  we  may  group  the 
terms  in  any  other  way  we  please. 

This  is  done  frequently  in  everyday  life.  Thus,  if  a  niercljaut 
owns  three  stores,  as  in  the  example  on  p.  48,  he  may  count  his  liabili- 
ties and  his  assets  for  all  these  stores,  or  he  may  count  them  separately 
for  each  store.  His  total  wealth  will  he  the  result  of  adding  all  these, 
and  will  be  the  same  by  either  method  of  working.  This  fact  is  often 
used  to  check  the  work. 

Ex.  1.  To  calculate  44  -f  36  -b  17  +  33  -f-  57  we  may  add  all 
together  or  we  may  group  in  parentheses,  thus : 

(41  +  80)  +  (17  +  .33)  +  57  =  80  +  -50  -f  57  =  187, 
or,  (44  +  36)  +  17  +  (33  +  57)  =  80  -|- 17  +  90  =  187, 

or,  (44  +  33)  +  3()  +  (17  +  57)  =  77  +  36  +  74  =  187. 

I  Which  of  these  is  the  easiest  way  ? 


48 


ADDITION   AND   SUBTRACTION 


[Ch.  Ill 


Ex.  2.  A  merchant  has  three  stores.  These  have  the  follow- 
ing amounts ;  the  debts  due  and  mortgages  to  be  counted  as 
negative  : 


Cash 

Fixtures 

Stock 

Acc'ts  Due 

Debts  Due 

Mortgage 

Total 

1st  Store 

$327 

$250 

$1200 

$463 

$651 

$    500 

2d  Store 

.f    56 

$615 

$2100 

$590 

$980 

$2000 

3d  Store 

$125 

$357 

$1570 

$   25 

$415 

$1400 

Total 

Find  (a)  the  total  value  of  each  stoi'e ;  (ft)  the  sum  of  these  three 
to  get  the  merchant's  wealth ;  (c)  the  total  cash  in  the  three  stores 
and  the  total  of  each  of  the  other  columns;  (d)  the  sum  of  these 
totals  to  get  the  merchant's  wealth.     Do  your  total  answers  agree '! 

EXERCISES   V:   CHAPTER   III 

Note.  The  student  should  check  all  answers  in  which  letters  are 
used,  by  substituting  numbers  for  the  letters  at  random,  as  suggested 
above. 

1.  The  gains  of  a  merchant's  two  stores  are  as  follows  for  a 
period  of  five  years.     Fill  in  the  spaces  left  blank : 


First  Store 

Second  Store 

Total 

A  VBI!A(iE 

1902 

11525 

-$7530 

1903 

$  1035 

-  f  5000 

1901 

$    105 

$   565 

1905 

-  $   355 

f  3325 

1906 

$   870 

$  9565 

Total  for  five  years 

Average  per  year 

:;()-:'3l]     GENERAL  RULES  — PARENTHESES        49 

Perform  mentally  the  following  additions,  abbreviating  by 
grouping  terms  as  often  as  possible  : 

2.  6  +  8  +  4  +  3  +  2. 

3.  12+17  +  28+35  +  30  +  5. 

4.  11  d  +  6d +  9(1 +  3  d  +  d  +  Ad. 

5.  17  xy  +  Sxy  +  12xy-\-23xy. 

6.  3  zH +  9  z't +  12  zH  +  ^t +  7  zH. 

Collect  into  single  terms  : 

7.  (12  be  +  9  be  +  8  be)  +  {7  be -3  be +  2  be). 

8.  (_  i^v  -  8  rv  +  13  r-v)  +  8  7-'i;  +  (7  i^v  -  12  r^v). 

10.  Average  the  expressions: 

(7  pq  -bpq+  pq),  (8  pq  -  3  pq  -  9  j)q),  (7  p)q  +pq-3pq). 

11.  Average  the  expressions  :  2  Imn  —  5  Imn  —  8  Imn, 
—  9  linn  +  ban  —  2  Imn,  6  Imn  +  4  /m/i  —  Imn. 

31.  Addition  of  Longer  Expressions.  We  may  now  add 
longer  expressions  by  grouping  together  those  terms  which 
are  "similar"  (or  "  like  ")  (see  §  8,  p.  8). 

Ex.  1.  One  farmer  has  7  horses,  12  cows,  and  4  sheep,  another 
farmer  has  6  horses,  5  sheep,  6  cows.  Find  their  total  posses- 
sions. 

We  say : 
(7  horses  +  12  cows  +  4  sheep)  +  (6  horses  +  5  sheep  +  G  cows)  = 
(7  horses  +  6  horses)  +  (12  cows  +  G  cows)  4-  (1  sheep  +  o  sheep)  = 
l:}  horses  +  18  cows  +  9  sheep,  but  we  do  not  try  to  add  cows  to 
sheep  or  to  horses. 

Likewise  (7  h  +  l2  c  +  4  s)  + (Qh  +  os  +  Q  c)  =  (7  h  +  Q  h) +  (V2  c+6c) 
+  (4  s  +  5s)  =  13  ^  +  18  c  +  9  s,  no  matter  what  h,  c,  and  s  mean. 


50  ADDITION   AND   SUBTRACTION  [Ch.  Ill 

Again  (6  ax  +  20  h^-x  +  5  aA)  +  (6  6%  +  30  ax  +  7  ah)  =  (6  ax  +  30  ax) 
+  (20  b'h:  +  6  b\c)  +  (5  ab  +  7  ab)  =  36  ax  +  26  h^x  +  12  ab,  uo  matter 
what  a,  b,  and  x  mean.  Check  this  addition  by  trying  «  =  2,  6  =  3, 
x=l.  Try  some  numbers  of  your  own  choice.  Notice  that  there  is 
no  attempt  to  add  togetlier  the  final  terms  3Q  ax  and  26  b'^x  and  V2ab, 
just  as  there  is  no  attempt  to  add  cows  to  horses  or  to  sheep  in  the 
first  example. 

Rule.  To  add  long  expressions,  lorite  them  in  columns 
with  the  similar  terms  in  the  same  column  ;  add  the  columns 
separately  ;  ivrite  the  total  sum  as  the  sum  of  these  separate 
results. 

This  was  done  in  Ex.  2,  p.  48.  Likewise  to  add  the  expressions 
(2  abc  +  3  a-x  —  7ab—4:bc  +  ~i  cx'^)  and  (6  ab  —  4  a-x  +  5  be  —  2  ex-  —  7) 
and  (4  ic  -  3  a6  -  3  cx2  +  2  a^x  +  4)  we  write 

Given  2  abc  +  8  a'^x  -  7  o6  -  4  be  +  5  cx^ 

-  4 a-x  +  Gab  +  5 be  -  2 ex-  -  7 

2  a^x  -  3  ab  +  4  ftc  -  3  cx^  +  4 

Sum  2  abc  +     a'^x  —  iab  +  5bc  —  3 

EXERCISES    VI  :  CHAPTER    III 

Add  the  expressions  given  in  each  of  the  following  exercises: 
[Check  each  result  by  substituting  random  numbers  for  the  letters.] 

1.  5a  +  7b-3c,  -12a  +  b-2c. 

2.  3x  —  4:y,7x  +  2y,  —2 x  —  y. 

3.  3.A--  -  4  (jy-  +  7  liz\  fx-  +  ijy-  +  //;i^  2fx-  +  rjy-  -  3  liz-- 

4.  al  —  <S  hm,  0  bni  —  2  en,  4  en  —  5  «/. 

5.  16  abc  —  19  a;//2;,  —  25  a?>c  —  1 1  .r//2,  15  xyz. 

6.  19 0^9 -I- 1 ; 5  hq  —  \5  cr,  — 17  bfj  +  rr  —  5  ap,  <S  />r/ — 11  cr  —  2  a/). 

7.  7  6c//2;  —  9  e<(zx  —  oa  bxy,  —  2  bcyz,         —  3  cazx  +  9  abxy, 
2  cazx  —  bcyz. 

8.  fKT^-S.r+T,  -5.T-4-2.r-3,  2.^■2-9a;-5. 

9.  x-  +  2  /i.i-  +  li-,  X-  —  /r,  /i-  —  hx  —  2  ar'. 


;jl-:5i>]  GENERAL    ULLKS  — rAllENTllESES  51 

10.  7  +  ;>  t  -  \  [It-,  -5^  +  3  (ji-,  G  -  2\  (jt-. 

11.  b{a  +  h)-- (i-+d)-\-\{x+!j),     -{a+h)-{c^d\ 
3(a  +  /y)  +  -'(o  +  d)-o{x  +  y). 

12.  3  (7 a;  -  5//)-  8(2x  -  3?/)+  10(.«  -  y\ 
-{1 X  -  r^ !l)  +  l{2x-'iy)-A{x- y), 
-(lx-oy)  +  2{2x--dy)-o{x-y). 

[Also  collect  your  result  iuto  ouly  two  terms.] 

13.  2  (-  X  -  3  //)  -o{x  +  3  y),  -  (-  x  -3y)  +  6  (x  +  3  y). 
[Collect  your  result  into  as  simple  a  form  as  po.ssible.] 

14.  (6  ax -^9  by -3  cz)  +  (6  ax  -i-9by-3cz) 

+  (6  ax  +  9  by  —  3  cz)  +  (G  ax  +  9  ?>//  -  3  cz). 

15.  Add  r^i.T  +  b^y,  ((.,x  +  '>o//,  ",.)■  + ^':'„'/,  1)y  collecting  the  coeffi- 
cients of  X  into  one  term  and  the  coefficients  of  y  into  one  term. 

The  use  of  lettera  ti-ilh  subscrijits,  as  here,  is  very  common;  thus,  a, 
(read  "  n,  sub-one  "),  a.„  a.,_,  b^,  b.,  (read  "  a,  sub-two,"  "  a,  sub-three,"  "  b, 
sul>-two,"  ''b,  sub-three,"  etc.),  are  very  often  used  in  the  place  of 
separate  letters.  They  should  be  treated  simply  as  separate  letters, 
used  to  indicate  separate  numbers,  and  no  meaning  should  be  attached 
to  the  small  figures  except  to  distinguish  a^  from  03,  for  example. 

32.  Subtraction  of  Longer  Expressions.  Just  as  before, 
to  subtract,  we  proceed  as  in  addition  after  changing  the 
sign  of  the  quantity  or  quantities  to  be  subtracted.  If 
there  are  several  quantities  to  he  suhtracted,  we  must  he  care- 
ful to  change  the  sign  of  each  one  before  proceeding  to  the 
addition  mentioned. 

Ex.  1.  A  firm  of  stockmen  have  $2500  cash,  600  sheep,  325 
cows,  a  mortgage  (debt)  of  $1500,  and  they  owe  25  horses  to 
another  firm.     What  is  their  total  wealth  ? 

The  total  wealth  is 

2.")()()  d  +  GOO  .s-  -t-  :?25  c  -  1500  d  -  25  h, 

where  the  letters  stand   for  the   article  having   the   initial. 


52  ADDITION   AND   SUBTRACTION  [Ch.  Ill 

Ex.  2.  If  one  partner,  wishing  to  set  up  an  independent  busi- 
ness, takes  part  of  the  debts  as  well  as  part  of  the  assets  :  $  500 
cash,  125  sheep,  150  cows,  a  mortgage  (debt)  of  $300,  what  is 
left  for  the  others  ? 

We  must  subtract  (500  d  +  125  s  +  150  c  -  300  rZ)  from  (2500  rf  + 
600  .s  +  325  c  —  1500  d  —  25^).  To  do  this  we  change  the  sign  of  each 
of  the  quantities  to  be  subtracted;  then  we  add  the  result  to  the 
original  amount. 

Original,  2500  d  +  600  s  +  325  c  -  1500  d  -25h 

Add  -  500  (/  -  125  s  -  150  c  +    300  d 

Answer,  2000  rf  +  475  s  +  175  c  -  1200  d-25h 

Ex.  3.    (4a-76x  +  3c-)-(6a  +  o?«-2c-). 

Original,  4  a  —    7  fe  +  3  c^ 

6a  +    5bx  —  2c^     (Change  signs  mentally  and  add.) 


A  nsiver,  —  2a  —  \2bx  +  bc^ 

This  holds  no  matter  what  a,  b,  x,  c  mean.     Try  it  if  «  =  4,  i  =  3, 
X  =  5,  c  =  1. 

EXERCISES    VII:  CHAPTER    III 
[Check  each  result  by  substituting  random  numbers  for  the  letters.] 
Subtract  the  second  expression  from  the  first : 
1.   a-b,2a  —  3b.  2.   31 —  5  n,  —I  — An. 

3.  6  x)j  —  4  a6,  —5xy-\-7  ab. 

4.  —'3kx-  —  i  bif  +  7  mz^,  —kx^  —  5  by-  +  2  mz^. 

5.  —  11  xy-  + 2  yz-,  —  10  xy- +  3  yz- +  zx-. 

6.  xyz  -i-  abc  +  hnn,  xyz  +  2  abc  +  3  Imn. 

7.  x  +  2  -y/xy  +  y,  x  —  2  Vx//  -f  ?/. 

8.  a'-2ab  +  b',b'-3ab-Aa-. 

9.  gryz  —  rpzx  —  pqxy,  rpzx  -\-  pg.r//  —  gryz. 

10.  cc-  —  ?/^  —  2  yz  —  z"^,  z^  +  x^  —  2  yz  +  y^. 

11.  2-\-llt-13(jf,l+12t-10gt-. 

12.  (Mi)x-3y)-2{x  +  y),    5(5  x -3 y) -3(x +  y).      Then 
collect  your  result  into  two  terms. 


:32-:j;5]  GENERAL   RULES  — PARENTHESES  53 

Perform  the  operations  indicated  : 

13.  (—9ax  +  7by  —  cz)-(-S ax  —  by +2  cz). 

14.  (2x-y)-{x-2y)  +  (-x  -  y). 

15.  (2x-y)-{x-2y)-(x  +  y). 

16.  (x'  -  2  hx  +  /r)  -  (x'  -  6  hx  +  9  h')  -  (x"  -  h')  +  (h'  -  hx). 

17.  (x' -2  xy^y-)-(3x'-5  xy  +  2y')  +  (2  X-  -3xy  +  f-). 

18.  Subtract  a^x  4-  azy  from  6i.^"  +  b-^  by  collecting  the 
coefficients  of  x  into  one  term,  and  the  coefficients  of  y  into 
one  term. 

33.    Removal  and   Insertion   of   Parentheses.     We   saw 

that  parentheses  preceded  by  the  sign  +  may  be  inserted 
or  renaoved  as  we  wish,  without  any  other  change. 

But  if  a  pair  of  parentheses  is  preceded  by  the  sign  — , 
the  whole  interior  is  to  be  subtracted  from  what  goes  before 
it.  We  must  therefore  cJiange  the  sign  of  each  term  inside 
the  parentheses  when  we  take  them  away. 

Thus,  (4  a  -  7  bx  +  :i c') - (6  a  +  ohx  -  2  c^)  =  i  a  -  7  hx  +  3  c^ 
—  Q  a  —  obx  +  2c^  =  —  2  a  —  12hx  +  5  c^,  which  is  the  same  as  the 
result  obtained  in  §  82.  Notice  that  the  Jirst  term  inside  a  pair  of 
parentheses  has  the  sign  plus  if  no  sign  is  written ;  in  removing  pa- 
rentheses preceded  by  the  sign  — ,  care  should  be  taken  not  to  over- 
look the  first  term. 

Similarly,  if  ive  insert  parentheses  preceded  hy  the  sign  — , 
we  must  change  the  sign  of  each  term  tve  inclose. 
Thus,  4  fi  -  7  6x  +  3  c-2  -  6  rt  -  5  /)x  +  2  c^  = 

(4a  -  6  a)  -  (7  bx  +  obx)  +  (3  c^  +  2  t--^)  ^ -2  a  -  12bx  +  o  c% 

which  is,  of  course,  the  same  result  obtained  before.  Notice  that  the 
middle  pair  of  parentheses  is  preceded  by  the  sign  —  ,and  that  the 
signs  of  the  terms  inside  of  it  are  changed  as  they  are  put  inside. 

To  remove   or  to  insert  parentheses,  change  the  sign  of 

1 7ione  1    ^j,.  ^j^^  terms  if  the  sign      _  [  precedes  the  paren- 
l  each  J  I       J 

theses.      (Read  "  none  "  with  -f- ;   read  "  each  "  with  -  .) 


54  ADDITION   AND   SUBTRACTION 

If  several  parentheses  are  used,  one  inside  the  other, 
remove  tlie  one  farthest  inside  lirst,  and  the  others  in 
their  order.  Tlie  student  should  read  §  11,  p.  10,  and 
sliould  state  the  forms  of  parentheses  used  in  algebra. 

After  considerable  pi-actice  the  student  should  be  expert  enough  to 
remove  any  one  pair  of  parentheses  without  touching  the  others ; 
he  may  then  find  it  more  convenient  not  to  use  the  order,  specified 
above. 

EXERCISES    VIII:  CHAPTER    III 

Remove  the  parentheses  in  the  following  exercises,  and  sim- 
plify as  much  as  possible : 

1.  7  +  (_3+2).  5.  -llb  +  [Sb-(2h-\-b)-3b'\. 

2.  7_(3_2).  6.  8  kz  -  [7  kz  -  {3  Jcz  -  5  Jcz)']. 

3.  _6  +  [5-(7+3)  +  12].      7.  8kz~[(7  kz-3kz) -5kz']. 

4.  13  _  [7  -  (2  -  5)].  8.  8  kz  -  (7  kz  -3kz~5  kz). 
9.  [Skz-7  kz^  -[3kz- 5  kz^. 

10.  [8  kz  -  (7kz  —  3kz)]  —  5  kz. 

11.  [G  mn-  -(8P- mrr-  +  3  n^  -  mn')  -  (22  m)r  -  8  P)]. 

12.  13  x>/z  —  [(5  abc  —  xyz)  —  (3  xjiz  -\-  7  abc  —  xyz)^. 

In  the  following,  remove  the  parentheses,  beginning  with 
the  outermost : 

13.  —  G  +  [o  —  (7 +3)  +  12].  Does  your  result  agree  with 
Ex.  3? 

14.  13  —  [7  —  (2  —  T))],     Does  your  result  agree  with  Ex.  4? 

15.  [6  mn-~{8  P-mn-  +  3  >P  -  mn'^)  -  (22  mn-  -8  P)].  Does 
your  result  agree  with  Ex.  11  ? 

16.  a-[-(--b)]. 

In  the  following  examples  rewrite  the  expressions,  inclosing 
the  last  two  terms  first  in  parentheses,  preceded  by  the  plus 
sign,  and  then  in  parentheses  })receded  by  the  minus  sign  : 

17.  a  +  b  +  c.  19.    (i  +  b  —  c.  21.    a—b+  c. 

18.  a  —  b  —  c.  20.    6  —  a  —  c.  22.    xtj  —  xz  +  Z'. 


PART   II.     APPLICATIONS.     LINEAR    EQUATIONS 

34.  Equations.     We  may  apply  the  knowledge  gained 
to  solve  many  problems  similar  to  those  of  Chapter  II. 

In  Chapter  II,  §  20,  we  had  an  equation 

c  =  ^n  +  200, 
where  c  means  the  total  cost  in  cents,  and  n  the  number  of 
copies  of  a  certain  printed  pamphlet.     If  the  cost  were  $4.25, 
or  425  cents,  we  could  find  the  number  of  copies. 

(1)  c=l  n  +  200, 
or,  since  c  =  425, 

(2)  425  =  in+  200. 
Subtract  200  from  each  side  of  the  equation  (2), 

(3)  225  =  I  n, 

since,  if  425  and  l  n  +  200  are  the  same  number,  it  follows  that  425 
less  200  is  the  same  number  as  i  n  +  200  less  200.  Now  multiply  each 
side  of  the  equation  (:})  by  2, 

450  =  n, 
since,  if  225  and  h  '*  are  the  same  number,  twice  one  of  them  is  the 
same  as  twice  the  other.     We  now  check  this  by  putting  n  =  450  in 
the  original  equation  (2)  ;  this  gives 

425  =  1  (-150)  +  200, 
which  is  evidently  correct. 

The  check  used  above  is  complete  (p.  5);  i.e.  the  check  leaves  no 
doubt  whatever  concerning  the  correctness  of  the  answer.  AVhenever 
the  answer  can  be  tried  directly  in  the  given  problem,  as  was  done 
above,  the  check  is  complete.  Hence,  this  should  be  done  whenever 
possible. 

35.  Operations  on  Equations.  The  work  just  done  is 
nothing  mure  than  we  could  hav^e  done  in  Chapter  II,  but  it 
is  carefully  stated.  In  particular,  we  consciously  subtracted 
200  fro)n  each  side  of  equation  (2),  and  we  multiplied 
each  side  of  equation  (3)  by  2.  The  student  will  see  that 
we  may  always  do  any  one  of  the  following  things  with- 
out disturbing  the  equality  : 

55 


50  ADDITION    AND   SUBTllACl'lON  [Cn.  Ill 

I.  We  may  add  the  same  number  to  each  side  of  an 
equation. 

II.  We  may  subtract  the  same  number  from  each  side 
of  an  equation. 

III.  We  may  multiply  each  side  of  an  equation  bj'  the 
same  number. 

IV.  We  may  divide  each  side  of  an  equation  by  the 
same  number,  wlien  tlie  division  is  possible  (see  p.  106). 

For  an  equation  indicates  that  the  two  sides  each  repre- 
sent the  same  number^  hence  the  result  of  adding^  subtract' 
ing,  multiplying^  or  dividing  by  the  same  number  on  both 
sides  must  be  the  same.*  Always  perform  the  operation 
upon  each  side  as  a  whole,  not  upon  a  part  of  it. 

36.  Definition  of  Linear  Equations.  An  equation 
with  one  unknown  letter  which  contains  the  unknown 
number  only  in  its  first  power,  in  the  form :  Ax  +  ^  =  0, 
where  A  and  B  are  known  numbers,  or  which  can  be 
reduced  to  that  form  by  the  operations  of  §  35,  is  called 
a  simple  or  linear  equation.  Most  of  the  equations  we 
have  studied  up  to  this  time  are  linear.   r§  20,  p.  25.) 

37.  Examples.  In  this  Chapter  we  have  learned  how 
to  remove  parentheses,  how  to  add  and  subtract  negative 
quantities  as  well  as  positive  quantities.  These  operations 
assist  in  solving  equations. 

Ex.  1.     Given  the  equation  2  .«  —  4  =  5,  where  x  is  an  un- 
known number  ;  to  find  x. 
(1)  2x-4  =  5. 

Add  4  to  each  side  :  2  x  =  5  +  4, 

or,  2x  =  9. 

*These  rules  really  state  the  axiom  that  there  can  be  only  one  result  for 
any  addition,  or  subtraction,  or  niulliplicati<in,  or  (possible)  division. 
Notice  that  each  of  tliese  operations  is  performed  by  means  of  a  number  ; 
if  any  expression  other  than  a  simple  number  is  used  in  any  of  them,  care 
is  necessary  to  insure  that  it  does  represent  a  number.     See  also  p.  10(5. 


3.)-;i7]  LINEAli    EQUATIONS  67 

Divide  each  side  by  2  :  x  =  |  =  4^. 

Check:  Put  U  for  x  in  (1) :  2  (4^  -  4  =  5,  or  9  -  4  =  5; 
since  this  is  correct,  we  see  that  the  vahie  of  x  found  is  correct. 

Ex.  2.    Given  the  equation 

(1)  2(;5..-4)-12  =  2[(a.--l)-(2x-3)], 
"where  x  is  unknown ;  to  find  x. 

Remove  the  parentheses  as  in  §  o3,  p.  53. 

(2)  6  X  -  8  -  12  =  2  (x  -  1  -  2  X  +  3), 
or, 

(3)  6x-20  =  2(-x+2), 
or, 

(4)  6x-20=-2x  +  4. 
Add  2  X  to  each  side : 

(5)  6  X  +  2  X  -  20  =  4. 
Add  20  to  each  side  :    6  x  +  2  x  =  4  +  20, 
or,  8  X  =  24. 
Divide  each  side  by  8  :  x  =  3. 

Check:  Put  3  in  the  place  of  x  in  (1), 

2(3.3-4)-12  =  2[(3-l)-(2.3-3)J, 

or,  2(5)-  12  =  2(2 -3), 

or,  10-12=2(-1), 

or,  -2  =-2, 

which  is  seen  to  be  correct;  the  answer  x  =  3  is  therefore  correct. 

The  general  plan  of  solution  is  as  follows  : 

(1)  Perform  all  indicated  operations.  If  necessary,  re- 
move fractional  coefficients  by  III,  §  35. 

(2)  Transpose  all  terms  that  contain  the  unknown  let- 
ter whose  value  is  required  to  one  side  ;  all  other  terms 
to  the  other  side. 

(3)  Combine  the  terms  on  each  side,  and  collect  the  co- 
efficient of  the  unknown  letter  (§  28). 

(4)  Divide  both  sides  by  the  coefficient  of  the  unknown 
letter. 

(5)  Check  the  answer  thus  found  by  trying  it  in  the 
(riven  equation. 


58  ADDITION    AND   SUBTRACTION  [Ch.  Ill 

EXERCISES   IX:  CHAPTER   III 

Solve  the  following  equations  to  find  the  value  of  the  un- 
known letter.    If  there  are  several,  solve  for  the  letter  specified. 

1.  x  +  3  =  T.  4.    ic  —  3  =  7.  7.   a;  +  3  =  —  7. 

2.  .r  -  3  =  -  7.      5.    -  .T  +  3  =  7.  8.    -  a;  -  3  =  —  7. 

3.  2.t;  +  5  =  17.      6.    9  a; -7  =  6.1;  + 8.      9.    2  .x-+5  =4a;- 1.^ 

10.  6a;  +  25  =  .^•-5.  12.    7  ?i -16  =  3  7i- 2. 

11.  6-5  z  =  3 -2.  13.    13p  +  50  =  25-(5  +  2p).K 

14.  12  a  +  1  -  (3  a  -  4)  =  2  a  +  8  +  (4  a  +  4).  t/ 

15.  6  A; +  9  =  2  A;- [4 -(10  +  2  A:)].  ■ 

16.  2.T  +  f  =  a;  +  2i.  19.    ^i^  x~  ^  =  2  x-l^. 

17.  ia:  +  i  =  ix  +  i.  20.    3(.T-5)-2(.T-4)=0. 

18.  9r-15  =  2r  +  3.  21.    2(7-2) +4(1  -  g)  =  6. 

22.  76^-139  =  196^  +  57. 

23.  2y-3-{-8{10-3y)  =  3-y.^  ^ 

24.  |(H  +  7)+i(n-4)=-i«. 

25.  cr  —  ax  =  ab.      [Solve  for  x  in  terms  of  a  and  A.] 

26.  i(,  +  2)-|(12-.)=i(.  +  9). 

27.  cz  —  c-  =  ac  +  be.      [Solve  for  z  ;  then  solve  for  a  ;  then  for  ft.] 

28.  Vt  =  s.      [Solve  for  v  ;  then  solve  for  s ;  then  for  <.] 

29.  ax  +  by  +c  =  0.      [Solve  for  x ;  also  solve  for  ?/.] 

30.  ax  —  by-\-c  =  a"—{by  —  c).      [Solve  for  x.] 

38.  Transposition.  In  any  equation  any  term  upon 
one  side  may  be  removed  from  that  side  by  subtracting 
it  from  both  sides.  This  lias  been  done  before.  The 
effect  is  the  same  as  if  we  simply  move  the  term  to  the 
other  side  of  the  equation  mid  change  its  sign. 

ThuSjin  Ex.  1,  §  87,  we  moved  4  from  the  left  side  of  equation  (1) 
to  the  right  side  with  its  sign  changed. 


;}7-:59]  LINEAR   EQUATIOXS  59 

In  Ex.  2,  §  37,  we  moved  —2x  from  the  right  of  equation  (4)  to 
the  left,  and  clianged  the  signi  from  —  to  + . 

This  operation  is  often  called  transposition ;  it  consists 
in  moving  the  term  to  the  other  side  of  the  equation  and 
changnuf  its  sign.  Hereafter  we  shall  use  it  freel}'.  The 
student  must  be  careful  to  transpose  only  terms,  not  factors. 

39.  Problems  stated  in  English.  Examples  stated  in 
English  may  be  put  into  algebraic  language  by  choosing 
convenient  letters  to  stand  for  quantities  mentioned  in  the 
example.      We  have  frequently  done  this  (Chapter  II). 

Ex.  1.  Tlie  sum  of  two  numbers  is  14,  their  difference  is  2. 
Find  the  numbers. 

Let  n  be  the  smaller  number ;  then  the  other  is  n  +  2,  since  the 
difference  is  2. 

Hence,  n  +  (n  +  2)  =  14, 

or,    .  2  n  +  2  =  14. 

Transjwse  2  :  2  n  =  12. 

Divide  by  2  :  n  =  6,  the  smaller  number. 

Tlie  larger  number  isn+2=6  +  2  =  8. 

Check:  the  numbers  being  6  and  8,  their  sum  is  6  +  8  =  14 ;  their 
difference  is  8  —  6  =  2. 

XoTE.  This  type  of  problem  often  arises  in  science  and  in  business. 
For  example,  a  boat  may  make  14  miles  per  hour  going  down  a 
stream,  but  only  2  miles  per  hour  going  up  the  same  stream.  In  that 
case  the  two  mimbers  in  our  problem  would  be  the  speed  of  the  boat 
in  still  water  and  the  speed  of  the  water  in  the  stream  ;  their  sum  is 
14  (in  miles  per  hour),  their  difference  ls  2  (in  miles  per  hour).  The 
work  just  done  shows  that  the  speed  of  the  boat  is  8  miles  per  hour, 
the  speed  of  the  stream  6  miles  per  hour. 

Ex.  2.  Two  men  enter  a  partnership,  the  first  giving  three 
times  as  much  as  the  second.  If  they  earn  $1000,  what  part 
should  each  receive  ? 

Let  X  =  the  share  of  the  first  partner,  in  dollars.  Then  1000  -  x  = 
the  share  of  the  second  partner,  in  dollars.  Since  the  first  should 
receive  three  times  what  the  second  receives, 

X  =  3  (1000  -  x),  or,  X  =  3000  -  3  x. 


60  ADDITION    AND   SUBTRACTION  [Ch.  Ill 

Transposing  3  ar,  we  get 

ix  =  3000. 

Dividing  by  4,  we  get 

X  =  750,  the  share  of  the  first  partner,  in  dollars. 
Hence,        1000  —  x  =  250,  the  share  of  the  second  partner,  in  dollars. 

Check:  $750  =  3  x  !$2.50;    $750  +  $2.30  =  -flOOO. 

Ex.  3.  If  two  grades  of  coffee  costing  25^  and  3")^  are  to 
be  mixed  so  as  to  sell  for  40^  a  pound  at  a  profit  of  25  %,  what 
parts  should  be  taken  to  make  50  lb.  of  the  mixture  ? 

Let  X  =  number  of  pounds  of  25  j?-  coffee. 

Then,  50—  a:  =  number  of  pounds  of  35^  coffee. 

Then  the  total  cost  of  the  50  lb.  is 

25  a;  +  35  (50  —  x),  in  cents. 

At  a  profit  of  25%  the  total  selling  price  of  50  lb.  is 
i^f  [25x  +  35(.50-x)]. 

If  this  is  to  sell  at  40^'  a  pound,  the  total  received  is  40  x  50  =  2000 
(in  (?).    Hence,     ^^,  ^..-^  ^  ^  3-  ^-,)  _  ^.^-j  ^  ^000, 

or,  I  [25  X  +  35  (50  -  .r)]  =  2000. 

Multiply  both  sides  by  4  : 

5  [25  X  +  35  (50  -  :f)]  =  8000. 
Divide  both  sides  by  5  : 

[25  X  +  35  (50  -  x)]  =  1600, 
or,  25  X  +  1750  -  35  x  =  1600, 

or,  1750  -  10  ,r  =  1600. 

Transpose  1600  and  10  a; :  150  =  10  x. 

Divide  by  10:  15  =  x. 

We  have  therefore  15  lb.  of  the  25  (^  coffee,  and  50  —  15  =  35  lb. 
of  the  35  J?  coffee. 

Check:  The  cost  of  the  mixture  is  15  •  25  +  35  •  35  =  1600  (in 
cents).     The  selling  price  is  therefore 

1600  +  25%  of  1600  =  1600  +  400  =  2000  (in  cents). 

2000  ^  50  =  40  (in  cents)  is  the  selling  price  per  pound.     This  is' 
as  desired ;  hence  the  answers  found  are  correct. 


39]  LINEAR    EQUATIONS  61 


EXERCISES   X:    CHAPTER   III 

The  student  will  best  proceed  as  follows  : 

(1)  Read  the  problem  carefully. 

(2)  Select  one  unknown  quantity,  and  denote  it  by  some  letter. 

(3)  Express  all  unknown  quantities  in  the  problem  in  terms  of  the 
one  just  selected. 

(4)  State  the  fact  given  in  the  problem  as  an  equation. 

(5)  Solve  the  equation  by  the  method  of  §  '47. 

(6)  Check  the  answer  by  trying  it  in  the  given  problem. 

1.  Divide  12(3  into  two  parts,  one  of  which  is  twice  as  great 
as  the  other, 

2.  Divide  32  into  two  parts,  one  of  which  is  three  times  as 
great  as  the  other. 

3.  Divide  75  into  two  parts  which  are  to  each  other  as  2  to  3. 

4.  One  partner  has  three  times  as  much  invested  in  busi- 
ness as  the  other.     How  should  a  profit  of  $  7000  be  divided  ? 

5.  Two  partners  have  invested  in  an  enterprise  $3000  and 
$  2000.     How  should  a  profit  of  $  360  be  divided  ? 

6.  Three  men  engage  in  business;  A  contributes  $1000, 
B  $800,  and  C  $750.  A  profit  of  $255  is  realized.  How 
should  it  be  divided  ? 

7.  The  sum  of  two  numbers  is  18,  their  difference  is  6. 
What  are  the  numbers  ? 

8.  The  sum  of  two  numbers   is  9,  their  difference  is  15. 
,  What  are  the  numbers  ? 

I  9.  A  ship  sails  against  the  current  of  a  stream  at  5  miles 
1  per  hour,  with  the  current  at  20  miles  per  hour.  What  is  the 
j  speed  of  the  ship  ?  of  the  current  ? 

10.  The   sum   of  two  numbers   is  s,   their  difference  is  d. 
What  are  the  numbers  ? 

From  Ex.  10,  find  directly  the  two  numbers  whose 

11.  Sum  is  16  and  whose  difference  is  10. 

12.  Sum  is  32  and  whose  difference  is  —  4. 


62  ADDITION   AND   SUBTRACTION  [Ch.  Ill 

13.  Sum  is  17  and  whose  difference  is  15. 

14.  Sum  is  —  9  and  whose  difference  is  —  5.- 

15.  The  difference  between  a  number  of  two  digits  whose 
sum  is  8,  and  the  number  formed  by  reversing  the  digits  is  18. 
What  is  the  number  ? 

16.  Find  two  consecutive  integers  whose  sum  is  31. 

17.  Show  that  the  sum  of  two  consecutive  integers  is  an 
odd  integer.     If  this  sum  is  a,  what  are  the  integers  ? 

Using  Ex.  17,  find  two  consecutive  integers  whose  sum  is : 
18.   55.  19.    13.  20.   71.  21.   45. 

22.  How  can  a  merchant  mix  10  pounds  of  tea  worth  31 
cents  a  pound  out  of  tea  worth  30  and  35  cents  a  pound  ? 

23.  A  merchant  mixes  two  grades  of  vinegar  which  cost 
him  55  cents  a  gallon  and  65  cents  a  gallon,  respectively. 
How  much  of  each  must  he  take  to  make  a  100-gallon  mixture 
which  he  can  sell  at  75  cents  a  gallon  with  a  profit  of  20  %  ? 

24.  Find  three  consecutive  integers  whose  sum  is  54. 

25.  Show  that  the  sum  of  three  consecutive  integers  is 
divisible  by  3.     If  this  sum  is  a,  what  are  the  integers  ? 

26.  What  is  the  amount,  if  $  200  is  lent  at  5%  simple  interest 
for  t  years  ?     In  how  many  years  will  the  amount  be  $250  ? 

27.  What  is  the  amount  if  $d  is  lent  at  5%  interest  for  t 
years  ?     When  will  the  principal  be  doubled  ? 

28.  Solve  example  27  for  a  rate  of  p%.  (The  answer  will, 
of  course,  be  in  terms  of  the  letters  p,  d,  t.) 

29.  At  what  rate  of  simple  interest  will  a  sum  of  money  be 
doubled  in  25  years  ?  in  t  years  ? 

30.  Find  three  consecutive  even  integers  whose  sum  is  24. 

31.  Can  you  mention  a  number  by  which  the  sum  of  three 
consecutive  even  integers  is  divisible  ?  Ans.  2,  3,  or  6.  (Stu- 
dent give  the  proof.) 

32.  What  must  be  the  sale  price  of  a  piece  of  land  in  order 
that,  after  deducting  an  agent's  commission  of  4%,  the  seller 
may  receive  $1200  ? 


39] 


REVIEW  68 


REVIEW  EXERCISES   XI:    CHAPTER   III 


Perform  the  operations  indicated  in  the  following  exercises, 
1_6^  regarding  them  first  as.  additions,  then  as  subtractions  : 

1.        6  ax -9  by  4.      x'-{-2xy  +  y' 

-  2  a.7;  -  3  by  2  x'  +    xy  -  7/ 


5.   n*  +  2n-  +  l 

-  5  xhf  +  3  xy  71*  -  2  71-  +  1 

—  9  x^y-  —  x'y"^ 


.  +  TVf 


6.    2 

Ix  —  my  ,  „    Ibz 

mx-\-ly  ^  t 


7.  How  is  the  average  of  two  numbers  x  and  y  found  ?  If 
we  know  the  average  of  x  and  y,  how  may  we  find  the  sum  of 
X  and  y ?  If  the  average  of  x  and  y  is  5,  what  is  x  +  y?  Plot 
the  figure,  showing  all  points,  the  average  of  whose  coordinates 
is  5. 

8.  Plot  the  figure  of  all  points  such  that  the  average  of  x 
and  y  is  —  3. 

9.  The  average  of  —x  and  y  is  —5.     Plot  the  figure. 

10.  The  average  of  x  and  —  3  >/  is  2.     Plot  the  figure. 

11.  The  average  of  y  and  2  is  x.     Plot  the  figure. 

12.  The  average  of  —5y  and  10  is  x.     Plot  the  figure. 

13.  Denote  by  y  the  average  of  x  and  —10  for  various  values 
of  X.  What  equation  connects  x  and  y?  Make  a  table  of  values 
of  x  and  ?/,  and  plot  the  figure. 

14.  Denote  by  y  the  average  of  —  3  a;  and  6.    Plot  the  figure. 

15.  Denote  by  ?/  the  average  of  —  2  x  and  —  3.  Plot  the 
figure. 


(34  ADDITION    AND   SUBTRACTION  [Cii.  Ill 

Perform  the  operations  indicated : 

16.  (ax  -  2  by  +  3cz)  +  (by  -  2cz-\-S  ax)  +  (cz  -  2  ax,  +  3  by). 

17.  (6  xY  +  3  x^  -  7  xy')  -(Zxy'-^f-l  xhf  +  2  xY)  + 
(x^  -  xY  +  x'y)  -  (4  ar^  +  4  /)  +  (10  xy'  -  7  xhf). 

18.  (x"  +  /  +  ^- -  2 .v^  -  2  ;^.c -  2  xy) - [(y-  +  z--2yz)+  (z"  + 
X--2  zx)  +  \x'  +  if  -  2  a-?/)  -  {x-  +  ?/-  4-  ^')]. 

19.  (.^-  _^  ^3  ^  ^3  _  3  ,p_y^^)    _    [-(_,^3  _  3  ^^2^  ^  3  ^^^2  _  23y  ^    (^^3  _ 

3  2;'a;  +  3  zx^  -  ar*)  +  (ar'  -  3  ;r?/  +  3  xy-  -  y^)']  +  [  (3  y-z  -  3  v/a!^)  + 
(3  z'x  -  3  zx^  +  (3  a'2?/  -  3  xy'')^. 

Solve  the  following  equations  for  the  letters  denoting  un- 
known quantities : 

20.  2(a;-5)  +  3(6-x-)  =  0.  22.  l(z-5)  +  (10-z)=^z. 

21.  6p+4-j_)=197)  +  25-7jx        23.  7-[5  +  (3-2n)]=15. 

24.  (1  -  iv)  -  [tv-  -  (3  w  _  5)]  +  [3  -  (5  -  ^'r')]  =  0. 

25.  (6A:-9)  +  [5-(3A:-2)]=0.       26.  (2 .^•+7)-(5+.r)  = -3. 

27.  2  aa;  —  a^  =  2  a;.^.      (Solve  for  a ;  then  also  solve  for  y.) 

Simplify  the  expressions  : 

28.  [(6  xy-7  ab) -2(oxy-3  ab)]  +  [(5xy-3  ab)  -  2  (6  xy 
-  7  ab)]. 

29.  [2  (10  2r  +  OjK/  -  r/)  -f  3  (2  //  -  3pq  +  2  r/)]  -  [5  (10  jr  + 
9  ^,g_  g2^_  (2  2>2_3 7>9  +  2  r)]  +  [2  (lO^r  +  9pq  -  <f)  -  3(2//-  - 
3pq  +  2q')]. 

30.  The  sum  of  two  numbers  is  3  a  +  b;  their  difference  is 
a  —  b;  what  are  the  numbers? 

31.  The  sum  of  two  consecutive  integers  is  4  A'  — 3;  what 
are  the  numbers  ? 

32.  How  many  pounds  each  of  spice  worth  20  and  50  cents 
a  pound  should  be  taken  to  form  12  pounds  of  a  mixture  worth 
30  cents  a  pound  ? 

33.  What  is  the  amount  at  compound  interest  at  rate  r  on 
principal  j)  for  two  years  ?  What  principal  will  yield  $005 
at  10  per  cent  at  compound  interest  for  2  years? 


39]  Sr.M.MAKY  ,  Go 


SUMMARY  OF  CHAPTER  III:   ADDITION  AND  SUBTRACTION; 
SIMPLE    EQUATIONS,  pp.  35-65 

Paht  I.     General  Rules  for  Operation;  Parentheses. 

pp. 84-54. 

Extension   of  the    Operations :    Addition   of   a   positive,   forward 
motion ;  subtraction  of  a  positive,  backward  motion. 

§  1>;5,  p.  84. 

FumlamentaJ  properties  of  addition  and  multiplication  :  five  rules: 
I.    a  -\-h  =  })  -\-  a  (Commutative  Law  of  Addition)  ; 
II.    a  X  l>  =  h  X  a  (Commutative  Law  of  Multiplication); 
TIL    a+  (b  +  c)  =  (a  +  /;)  +  c  (Associative  Law  of  Addition)  ; 
IV.    a  X  (b  xc)  =  ((t  X  b)  X  c  (Associative  Law  of  Multiplication)  ; 
Y.    a(b  +  c)  =  ab+  ac  (Distributive  Law). 

These  five  are  axioms.     J^xerclses  I.  §  24,  pp.  34-86. 
To  add  a  neyatire  nuinJter  :  backward  motion. 
To  subtract  a  negative  numlier:  J7)rinard  motion. 

'{Adding         \  f  subtract  in;/ ] 

\             '           >   a    neaatire   number:    equivalence  to   -^      ,  ,.              \    ^ 

[  Subtracting  )             '  I  adding         J 

positive  number  of  the  same  amount. 
To  subtract  any  number  :   change  its  sign  and  add.  §  2'),  pp.  86-37. 

To  add  several  nundiers  :   the  difference  of  negative  total  and  positive 

total  amounts  with  sign  of   greater   total   amounts.    Exercises  II. 

§  26,  pp.  37-39. 
E(iuiralent   Expressions:    «  +(-  b)   and  a-b;    a  -  {- b)  and  a  +  b. 

Exercises  III.  §  27,  pp.  89-44. 

To  add  similar  monomial  terms  :  sum  of  coefficients  x  common  factoi- ; 

to  subtract:  change  sign  and  add.     Exercises  IV. 

§§  28,  29,  pp.  44-45. 
Use  of  parentheses :  grouping  of  terms  in  addition.     Exercises  V. 

§  80,  pp.  ■I7-l!». 
To  add  longer  expressions:  add  similar  terms  in  columns.     Exercises 

VI.  ■  §81,pp.49-.-)l. 

To  subtract  a  longer  expression  :  change  the  sign  of  each  term  and  add. 

Exercises  VII.  §  82,  pp.  51-53. 

r  no  1       .  .^     f  +  1 

To  remore  or  insert  parentheses  :  change    \         \    signs  it    \         >   pre- 
cedes the  parentheses.     Exercises  VIII.  §  83,  pp.  53-54. 


60  ^     ADDITION   AND   SUBTRACTION 

Part  II.     Applications  ;  Linear  Equations.  pp.  55-64. 

Equations  :  typical  solution ;  complete  check.  §  -M,  p.  55, 

Permissible  Operations:    add,  multiply,  divide,  subtract ;    operate  on 
each  side  as  a  ivhole.  §  :i5,  pp.  55-56. 

Linear  or  Simple  Equations :  first  power  of  unknown.    §  o6,  p.  56. 

Solution  of  Examples:  typical  solutions;  plan  of  solution.     Ex- 
ercises IX.  §  ;^7,  pp.  56-58. 

Transposition:  change  of  sign  ;  tran^tpose  terms,  not  factors. 

§  :58,  pp.  58-59- , 

Problems  slated  in  English :  typical  solutions ;  directions  for  student. 
Hxercises  X.  §  39,  pp.  58-62. 

Review  Exercises  XI.  pp-  63-64. 


CHAPTER    IV.      MULTIPLICATION   AND    DIVI- 
SION;   FACTORING;    APPLICATIONS 

PART    I.     MULTIPLICATION    AND    DIVISION    OF    NUM- 
BERS   AND    MONOMIALS 

40.  Multiplication.  As  in  the  case  of  addition,  we  wish 
to  extend  the  idea  of  multiplication  so  that  we  can  mul- 
tiply any  kinds  of  numbers  together.  We  shall  always 
keep  the  rules  given  on  p.  35,  and  in  every  other  way  we 
shall  try  to  preserve  the  spirit  of  what  was  known  as  mul- 
tiplication in  elementary  arithmetic. 

Multiplication  was  originally  applied  to  integers  only. 
In  multiplying  5  by  4  we  take  5  four  times  and  add ;  thus, 
5  X  4  =  5  +  5 +5  +  5  =  20. 

This  notion  very  clearly  does  not  hold  for  fractions. 
In  multiplying  12  by  |  we  do  7iot  take  12  two  thirds  times, 
for  that  is  absurd. 

To  multiply  12  by  |  we  may  note  that 

12xf  =  |x  12  =  1  +  1  +  .-.  (12  times)  =8, 

by  virtue  of  our  Rule  II,  p.  35.      In  general,  if  a,  ^,  c,  are 

any  positive  integers, 

h      h  h  ,  h  ,        ,     ,.        ^      ah 

ax-  =  -xa  =  -H [-•••(«  times)  =  — 

c      c  c      c  c 

If  we  now  take,    say  f  x  |^,  we  note  that 
(f  X  f )  X  35  =  f  X  (I  X  35)=  f  X  20  =  12,    by  IV,  p.  35. 

Hence,  (fx|)x  35  =  12, 

or,  by  IV,  p.  56,  |x-4=^2. 

07 


68  MULTIPLICATION   AND  DIVISION  [Ch.  IV 

Likewise,  in  general, 

for    (-  X-]  X  bd  —  -  X  l-  X  M] 
\h     dl  b      \d  } 


a      c  _ac 
l  ^  d~bd.' 


"        1 

=  ~  X  CO. 

b 


Thus,we  arrive  at  the  usual  rule  for  multiplying  fractions  by  means 
of  the  Rules  II  and  IV,  p.  85. 

The  elementary  definition  of  multiplication  (for  in- 
tegers) is  :  The  product  of  two  integers  is  found  hy  taking 
the  multiplicand  as  ma7iy  times  as  is  indicated  hy  the  multi- 
plier and  adding.  We  extend  this  definition  for  other 
numbers  by  saying  :  The  product  of  any  two  numbers  shall 
be  such  that  the  rides  of  p.  35  shall  remain  true. 

Thus, the  rule  for  fractions  results  in  this  way,  as  shown 
above.  We  shall  use  this  principle  to  find  the  product  in 
all  new  cases. 

As  another  example,  consider  the  product  of  any  num- 
ber and  zero ;  say  4x0.     We  say 

0x4  =  0  +  0  +  0  +  0  =  0. 

But,  4x0  =  0x4  by  Rule  II ;  hence,  4  x  0  =  0. 

TJie  product  of  any  number  and  zero  is  zero. 

The  proof  shows  this  to  be  true  for  any  integer  only.  The  follow- 
ing argument  shows  that  it  is  true  also  for  any  fraction : 

To  find  Ox-;  consider  (c  —  r)x-=Ox-' 
b  ^         '      b  b 

Ox-  =  (c  —  c)~  = =  0. 

b      ^         U,        b  b 

This  proof  need  not  be  learned  at  this  time. 

41.  Products  of  Negatives  and  Positives.  It  frequently 
becomes  necessary  to  multiply  a  negative  number  by  a 
positive  number. 


4(1-41]  NUMBERS    AND   MUXO.MIALS  G9 

Thus,  if  a  man  makes  ten  debts  of  i50  each,  his  total 
debt  is  $500,  i.e.  (-50d)xl0=-  500  d.  This  holds 
whether  d  means  dollars  or  anything  else. 

In  general,  (—  a)  x  5  =  —  rt6. 

To  iimltiply  a  negatioe  quantity  hy  a  positive  quantity., 
multiply  tlte  two  amounts  as  if  both  were  positive  and  prefix 
the  sign  —  to  the  answer. 

Tliis  is  convenient  also  in  solving  equations.  Thus,  the  equation 
.S  X  —  6  =  0  gives  X  —  2;  check  :  3(2)  —6=0  (correct).  But  if  we 
first  transpose  '6x,  we  get  —  6  =  —  3a;;  an  attempt  to  put  2  in  place 
of  X  here  gives  —6  =  —  3(2),  which  must  be  correct  in  order  to 
avoid  contradicting  the  previous  check. 

Since  the   factors  should  give   the  same    result  when 
multiplied   in  any  order,  the  same  rule  holds  for  multiply- 
ing a  positive  number  by  a  negative  number.      In  formula : 
a  x{—by  =  —ab. 

We  can  easily  prove  that  these  results  must  be  true  if  the  rules  on 
p.  35  hold  true,  but  it  is  not  necessary  for  the  student  to  learn  tliis 
proof  at  this  time. 

For  o  X  [A  +  ( -  6)]  =  a  X  (6)  +  a  X  (-  i),  by  Rule  V. 

But,  a  X  [i  +  (-  6)]  =  0  since  b  -1  =  0. 

Hence,  ah  +  a(—  1)=  0, 

or,  a  X  (  -  ft)  =  —  afi.  which  is  one  of  our  rules. 

Moreover,  (~  !>)  x  a  =  a  x  (—  b)  by  Rule  II,  —  —  ab,  which  Ls  the 
same  as  our  rule  above. 

EXERCISES    I:    CHAPTER    IV 

1.  A  man  owes  twelve  debts  of  $30  each.  What  is  his 
total  indebtedness?  State  the  problem  in  terms  of  negative 
numbers. 

2.  Each  inhabitant  of  a  city  pays  a  certain  tax  of  $10. 
What  is  the  increase  in  the  wealth  of  the  city  due  to  the  de- 
parture of  fifteen  residents ".' 

3.  How  much  is  the  downward  pressure  of  a  bar  of  iron 
lessened  by  the  attachment  of  twelve  springs,  each  pulling  up 


70  MULTIPLICATION   AND   DIVISION  [Ch.   IV 

with  a  force  of   75  pounds  ?     Express  in  terms  of   negative 
numbers. 

Perform  the  following  multiplications  : 

4.  5x-2.  9.       3x(-lx7).    14.      2x-(3a;x2|) 

5.  7x-3.  10.       7x-15(L        15.      Ox -3.* 

6.  4  X  -  1.  11.       G  X  -  13  A-.        16.  -3  X  0. 

7.  _  4  X  1.  12.       6  X  -5  X  9  kl.   17.  -21  X  0  X  3  x. 

8.  -  2  X  5.  13.   -  9  X  3  x-yz^.         18.  -  7.2  x  2.3  x-y. 
19.   -5x6(6c+co  +  a6).  20.    -^ix-+y'-2xy)x\0. 

21.  _i3r^^  +  ^:^  +  ^xo. 

\q-r      r-p     p-qj 

22.  In  arithmetic  we  multiply  by  taking  the  multiplier  as 
many  times  as  the  number  of  units  in  the  multiplicand  and 
adding.  Supposing  this  to  be  a  proper  method  whenever  the 
multiplier  is  a  positive  integer,  show  that  —  5  x  12  =  —  60 ; 
0x7=0. 

42.  Negatives  multiplied.  Tt  is  frequently  necessary  in 
algebra  to  multiply  one  negative  number  by  another.  It  is 
convenient,  and,  in  fact,  necessary  if  we  are  to  avoid 
contradictions,  to  say  that  the  product  is  positive. 

Thus,  given  the  equation  3  x  +  6  =  0,  we  find  x  =  -  2 ;  check : 
3(-2)  +  6  =  0  (correct).  If  we  first  transpose  'i u:,  we  have  6  =  -3x; 
an  attempt  to  put  -  2  in  place  of  x  in  this  form  of  the  equation  gives 
6  =-  3(-  2),  which  nuist  be  correct  if  we  are  to  avoid  contradict- 
ing the  previous  check.     Hence, we  say  (—  3)(—  2)=  6. 

In  general,  ( -  «)  X  (—  J)  =  +  ab,  or, 

To  midtiply  one  negative  number  by  another,  midtiply 
their  amounts  as  if  both  tvere  positive  and  prefix  the  sign  -)~ 
to  the  answer. 


*  Observe  that  +0  =  —  0,  since  each  means  no  motion  at  all  on  our 
scale.  Hence,  in  the  answers,  write  merely  0  whenever  either  +  0  or  —  0 
would  result. 


41-i;!]  NUMBERS    AND   iMOXOMIALS  71 

This  is  easily  proved  by  the  rules  of  p.  35.     For 

(_„)x(i-i)  =  (-a)x  b+(-a)(-b), 
,,r,  0  =  -  ab+(-  a)(-b), 

nr.  +„^,  =  (_  „)(-/')• 

[The  student  need  not  learn  this  proof  at  this  timtf.] 

The  preceding  rules  may  be  stated  as  ftdlows: 

—  a  X  +  b  gives  —  ab^ 

+  a  X  —  b  gives  —  ab, 
f  —  a  X  —  b  gives  +  a^, 

and,  of  course,         +  a  x  +  b  gives  +  ab,  or, 

Rule  of  Signs  :     In  multiplying^  like  signs  give  +  and 
unlike  signs  give  —. 

EXERCISES    II  :  CHAPTER   IV 

Perforin  the  following  multiplications  : 

1.  -  2  X  -  3.         5.      -  7  X  -  5.  9.  - 13  X  -  5  rf. 

2.  —  3  X  7.3.  6.  5  X  —  7.  10.  —    4  X  —  7  xyz. 

3.  7  X  —  5.         7.    —    Ix— 1.  11.         7x— 5x— .2, 

4.  —  7  X  o.  8.    —  13  x  0.  12.  —    1  x  —  1  X  —  1. 
13.        2  X  - 3 hhjz  X  -  13.       14.    —(jkzx-3x- 5. 

15.    -  i  X  i  f  (^/-  +  li'x  +  xif) .     16.     -  7  .r-j/  X  0  X  -  3|. 

9  ^\qs+rtj  om^—ii^ 

19.     -lX-lx-lX-1.       20.     -IX -IX— IX-lX -1. 
21.     (-3/.  22.     (-3)^  23.    (-3)^ 

43.    Multiplication  of  Monomials  by  Rearrangement.    Mo- 
nomials mag  be  multiplied  by  rearranging  their  factors. 

Thus,  i  ub  X  'i  ah  =  (i  X  a  X  />)  X  Qi  X  a  x  b) 

=  (1  X  :5)  X  (a  X  a)  X  (b  x  b) 
=  12  X  o-  X  b'\ 
since,  by  Rule  II,  p.  85,  the  order  of  multiplication  is  immaterial. 


72  MULTIPLICATION    AND   DIVISION  [Ch.  IV 

Check:  If  we  put  a  =  2,  h  —  Z,  we  find 

4  .  2  .  3  X  3  •  2  •  3  =  12  •  2^ .  32  =  12  •  4  •  9  =  432  (correct). 

Likewise,  ( -  4  ah)  x  (3  ab)  =  -  (4  «/>)  (3  ul>)  =  -  12a2//2, 
by  the  rule  for  signs  given  just  above. 

Check:  '         If  we  put  (/ =  2,  ?y  =  3,  we  find 

(  _  4  .  o  .  o)  X  (3  .  2  •  3)  =  -  12  .  22 .  32  =  -  12  X  4  X  y  ==  -  432  (correct). 

EXERCISES    III:     CHAPTER    IV 

Perform  the  following  nmltiplifations  by  the  method  just 
given.  Check  each  result  by  svibstituting  numerical  values  for 
the  letters. 

3.  —12x-y  5.        1 2> 

—  3xjr  —  8  (^■  +  m  +  n) 

4.  9iiv^io  6.    —2(;x  +  y  +  z  —  l) 

-  7  xvhi'  -  3  (x  +  y  +  z-  ly 


1. 

-5a 

6b 

2. 

-2kr 

-3x 

7.  —  2  a  X  —  2  a  X  —  2  a.                8.    —  7  x-if  X  3  .w/2;  X  —  kz^. 

9.  (-3a')l                 10.    (2m-Hy-.                11.    (-3.}r»^. 

12.  -  (a  +  6)(a--  +  ?/-)■'  X  3{x-+  fjia  +  hf  x  -  2  (a  -{-b){x-  +  f). 

13.  l-m^ri*ifq^r  x  —  2  Pm'^n^i/fjr  X  3  l^iu-'v''p<fr  X  —  4  hn^n2r(fr\ 

44.  Multiplication  of  Simple  Powers.  By  the  same  rule 
we  may  multiply  any  simple  powers  (§  9,  p.  8)  of  the  same 
quantity. 

Thus,      22  X  23  =  (2  X  2)  X  (2  X  2  X  2)  =  2  X  2  X  2  X  2  X  2  =  2^, 
and,  a^  y.  a^  —  (a  x  a)  x  (a  x  (t  x  <i)  =  a  x  a  x  a  x  u  x  a  =  a^, 

no  matter  what  <i  may  mean.     Likewise, 

u'^  X  a^  —  a  ■  a  ■  (I-  a  x  a  ■  a  ■  a  =  (C. 

In  general,  for  simple  })o\vers, 

a"'  X  a"  =  a  ■  a  •■■  (m  times)  x  a  •  a  •■■  (^n  times) 
=  (I  ■  a  ■  a  ■  ■  •  (m  +  n  tinu's)=  «""'". 


43-1.-}]  NUMBERS   AND   MONOMIALS  73 

'I'his  rule  is  proved  now  only  for  simple  powers,  i.e.  when  w  and  n 
are  positive  iutegeis.  Later  we  shall  prove  that  it  holds  for  other 
values  of  )n  aud  n.     See  Chapters  Vil  and  XI. 

Note  that  the  exponent  in  the  product  is  the  sum  of  the 
(jiven  exponent.^. 

If  this  rule  is  forgotten,  it  will  speedily  be  found  again  by  actually 
writing  out  the  meaning  of  the  separate  factors,  as  above;  but  this 
rule  is  simpler.     It  is  well  to  remember  the  example  given  above, 

a^  X  «^  =  a^, 

in  order  to  remember  the  general  rule. 

Do  not  forget  that  a  quantity  with  no  exponent  expressed  really 
has  an  exponent  unity,  „  _  „i 

45.  Final  Rule;  Monomials.  By  using  the  above  rule 
we  may  multiply  more  simply. 

Thus,  (4  a^i2)  X  ( -  2  ah'')  =  (4.-2)  (a^  •  a)  (h^-  ■¥)  =  -S  aVA 

Do  not  fail  to  choose  the  proper  sign  for  the  product 
according  to  the  rule  of  signs  given  above. 

Several  factors  may  be  multiplied  together  at  once,  if 
care  is  used. 

Thus,  (4  a^h^)  X  (-  2  nh^)  x  (-3  ah)  =  -  24  a*b\ 

The  coefficient  in  the  product  is  the  product  of  the  coeffi- 
cients in  the  given  factors  ;  the  exponent  of  any  letter  in  the 
product  is  the  sum  of  the  exponents  of  that  letter  in  the  given 
factors. 

If  this  rule  is  forgotten,  the  same  problems  can  all  be  done  by  §  43. 

The  product  of  any  number  of  positive  factors  is  posi- 
tive. The  product  of  an  even  number  of  negative  fac- 
tors is  positive  ;  of  an  odd  number  of  negative  factors, 
negative.  The  product  of  any  number  of  positive  and 
negative  factors  is  positive  when  the  number  of  negative 
factors  is  even,  and  negative  when  the  number  of  negative 
factors  is  odd. 


74  MULTIPLICATION   AND   DIVISION  [Cii.  IV 


EXERCISES   IV:   CHAPTER   IV 

Perform  the  following  multiplications  by  the  method  just 
explained;  do  the  first  five  also  by  the  method  of  §  43.  Check 
each  result  by  substituting  numerical  values. 

1.  —  3  a^x  X  —  2  ax^y.  5.    —3bcx  —  cax—  ab. 

2.  -3zc'x2  zV.  6.    6  phfr'"  x  -  3  pry^  X  -  2  rp,,. 

3.  7gtx—^L  7.-4  ab-c  x  —  3  a-bc  x  —  2  abc-. 

4.  9(22  X  33)  X  3(2  x32).        8.  2.5  m-u"x  -8  mYx  +inV, 
9.  5/y/t3  X  -  3ff/%  X  -  2f-h\ 

10.  7.2  a'^bc-  X  -  4.1  ahh  x  1.3  a-bc\ 

11.  -3{y  -  z){z  -  X)  x2{z  -  x){x  -  II)-  X  -  {y  -  z). 

12.  -  2(-3  bc){-2  ca)  X  -3(-  2  ca)(-a/^)  x  (-  ab)(-  3  &c). 

13.  5^1^x-3{-y/hgfx-2{^hgf. 

Simplify  the  result  as  much  as  possible. 

15.    a™  X  a".  16.    -  cr  X  —  a".         17.    (-  ay  x  (-  «)". 

18.  (-a)2;  (-a)'^   (-a/;   (-«/;   (-«/;   (-a)^ 

19.  (—a)'",  if  m  is  even.  20.    (—a)'",  if  m  is  odd. 

21.    {-2x)\  22.    (3ri"¥)l  23.    {- 4:xhifi2  xy-y. 

46.  Division.  Division  is  the  reverse  of  multiplication  ; 
the  result  of  division  is  called  the  quotient ;  that  is, 
dividend  -;-  divisor  =  quotient,  if  quotient  x  divisor  =  divi- 
dend. 

For  example : 

12  -  .3  =  4,  for  4x3  =  12. 

Again,  1^1  =  ¥-,  for  V  x  §  =  |. 

T  1  (I      c      ail    e      (id      c      a 

In  general,  -  -4-  -  =  — ,  tor  — ^  x  -  =  -  • 

b      d      be  be       d      b 

We  shall  consider  this  again  later  (Chap.  V,  p.  1.34). 


45-16]  NU.MBEliS   AND   MONOMIALS  75 

A  negative  number  divided  by  a  positive  number  gives 
a  negative  result ;   thus, 

-  12  -  3  =  -  4,  for  -4x3  =  -  12. 

A  positive  number  divided  by  a  negative  number  gives 
a  negative  result ;  thus, 

(+12)-(-3)=-4,  for  -4  X -3  =  12. 

A  negative  number  divided  by  a  negative  number  gives 
a  positive  result ;   thus, 

(_12)--(-3)=  +  4,  for  4  x  -  3  =  - 12. 
The  rule  may  be  stated  in  short  as  follows : 

Rule  of  Signs.  In  division,  like  sipis  (jive  plus ;  un- 
like signs  give  minus. 

Notice  tliat  this  i-iile  reaih  precisehj  like  the  rule  for  signs  in  multi- 
plication, p.  71.     Let  the  student  show  why  this  should  be  the  case. 

Zero  divided  by  any  number,  not  zero,  gives  zero;  thus, 
0-3  =  0,  for  0x3  =  0  (p.  68). 

The  quotient  of  any  number  divided  by  zero  does  not 
exist ;  thus,  3  -r-  0  =  ?  (does  not  exist),  for  the  question, 
?  X  0  =  3,  has  no  answer. 

EXERCISES   V.    CHAPTER   IV 

Perform  the  following  divisions: 

1.  _18--G.    3.  18---G.    5.   -8 --2.    7.   -5-h2. 

2.  -18-6.        4.  15 --3.    6.  7 --3.        8.   -13 --5. 
9.  G  a  -!-  -  2  a.       10.  15  bcp  -=-  -  3.     11.   -  17  xyz  -=-  -  2  xyz 

12.    -  3  (be  +  ca  +  ah)  -.-  -  2.       i3.    -  ^'  -  -  |?. 

14.  IG^-^^-^V-^f"^^" 

\     Ew     J  \     Rw     , 

15.  —  abcvyz  -H  —  abc.     16.  0  — •  —  G  x-yz^.     17,    —  51  m  —  0. 


76  MLILTIPLICATIOX    AND   DIVISION  [Cii.  iV 

47.  Division  of  Simple  Powers.  We  may  find  out  how 
to  divide  uuuiy  other  expressions  by  the  same  reasoning, 
when  the  exponents  are  positive  integers. 

Thus,  a^  -i-  a^  =  a^,  because  a^  x  a^  =  a^ 

and  a'  ^  a"^  =  a^  because  a^  x  a^  =  a", 

and         a"'  h-  «"  =  a'"""  because  a'"'"  x  a"  =  a"". 

This  rule  is  proved  now  only  when  m  and  n  are  positive  integers, 
and  VI  is  greater  than  n.  Later  we  shall  prove  that  it  holds  for  other 
values  of  m  and  n.     See  Chapters  VII  and  XI. 

Hence,  in  dividing  powers  of  the  same  quantity,  the  ex- 
ponent in  the  quotient  is  the  difference  of  the  exponents, 
the  exponent  in  the  divisor  being  subtracted  from  that  in  ' 
the  dividend. 

The  rule  is  seen  also  by  writing  out  the  factors  in  full : 

a-!  ^  ^,^<LJLJilA:l^jtl  =  a  .a.a  =  a\ 

^  .  <ji  •  ji  ■  ji 

as  above.     If  the  exponent  of  the  divisor  is  the  greater,  the  quotient 
is  naturally  a  fraction.     Thus, 

a*  H-  a''  =  </i  ■  ji  ■  <ti  ■  fi ^       1       ^  1 . 

^  •  </i  •  ^  •  ft  •  a  •  a  •  a      a  ■  a  •  a      a'^ 

These  divisions  can  be  carried  out  always  by  carefully  writing  down 
the  meaning  of  each  factor,  as  in  this  example. 

It  should  be  noticed  that 

—  =  1  •  for  —  =  ^  '  ^  '  ^^  =  1. 
a^        '  o^      ^i  -(ji  •  ^ 

In  general,  the  quotient  of  two  identical  powers  of  the  same  quan- 
tity is  unity. 

48.  Division  of  Monomials.  To  divide  one  monomial  by 
anotlier  we  merely  apply  the  rules  above.      Tor  example : 

=  4  <r/t. 

8  aVy'a^ 


47-lSJ  LUMBERS    AM)   MONOMIALS  77 

This  may  also  bo  done  as  follows: 

The  cancellation  is  done  npon  the  same  principle  as  in  arithmetic: 
the  numerator  (dividend)  and  the  denominator  (divisor)  may  each 
be  divided  by  tlie  same  quantity  without  altering  the  value  of  the 
fraction  (quotient). 

As  a  convenient  rule  we  notice  that  the  coefficient  in  the 
quotient  is  the  quotient  o/ ^Ae  given  coejfjcients ;  the  exponent 
of  any  letter  in  the  quotient  is  the  difference  of  the  given 
exponents  of  that  letter  in  the  order  mentioned  above. 

Remember  the  rule  of  signs  :  the  quotient  is  positive  or  negative 
according  as  dividend  and  divisor  have  like  or  unlike  signs. 

EXERCISES   VI :  CHAPTER  IV 

Perform  the  following  divisions ;  check  each  i-esult  by  sub- 
stitution of  numerical  values : 

1.  —15  aire  ^. —  5  be.  9.  a"'b"cf^. a''6'c*. 

2.-5  xy  --  3  xf.  10.  ( -  1 )«---(- 1)^ 

3.  12  x'jf  --  -  4  xif.  11.  ( -  a)'--^  -  a\ 

4.  —  5  ab-x  -. ab-x.  12.  (—  af^a^. 

5.  3  a-lM  -=-  -  2  abc.  13.  -a^^  -  a\ 

6.  -  15  hnir  -  -  5  l-mn\  14.  0  --  -  5p-qh'^. 

7.  -Qkz'^:Mcrz.  15.  -5/A/V'-0. 

8.  -  1 7  Mv  -^  -  5  Jrlhr.  16.  (x  +  .v)'X«  +  b)-^(x  +  y)%a  +  b). 

Perform  the  indicated  operations,  expressing  the  results  in 
simplest  form : 

12  a-b  X  —  3  ab^  — 14  xf/z  x  3  .cV  x  — 10  yz^ 


-Qab  -  21  xYz^ 

iirp^  X  —  5  mp  or'?/™2"  X  bx"~' 

lO^nhv'  uIkk"'"')/' 

x-BWh"  22   (■r+y)Hri+by 

ABab         '  '               2{x  +  y)-{a+by 


—  13  m)rp^  X  —  5  mp      __      a.rhrz"  X  bx'"~y~"'z 

lo.      •      21. '- 

^^     A'a^"X-BWh"  22.    (■v+y)Hri  +  by  X  (x  +  yy(a-\-by 


PART  II.     MULTIPLICATION  AND  DIVISION  OF  LONGER 
EXPRESSIONS 

49.  Monomial  x  Binomial.     In  §  28,  p.  44,  we  saw  that 

and  so  on.     We   may  now  multiply  any  binomial    by  a 
monomial  by  the  same  principle. 

Thus,        (3  a2  +  2  ^2)4  ah  =  (3  a^~) (4  ah)  +  (2  h"^) (4  ah} 

=  12  a%  +  8  ah^. 

The  product  of  a  moyiomial  and  a  hinowial  is  the  sum  of 
the  products  of  the  7nonomial  times  each  of  the  terms  of  the 
binomial. 

This  is  i-eally  merely  a  restatement  of  Rule  V,  p.  35: 
a(b  +  c)  =  ab  +  ac. 

Care  should  be  taken  not  to  overlook  negative  signs. 

In  checking  such  problems,  it  is  often  sufficient  to  set  each  lettei 
equal    to    unity.      In   the   preceding   example,    if    a  =  \    and   6  =  1, 
(3  fl2  +  2  lfl)i  oft  =  (3  +  2)4  =  12  +  8  =  20,  and  12  a^l>  +  8  ab^  =  12  +  8 . 
=  20  (correct).     This  check  is  not  complete,  but  it  serves  to  convince 
us  that  the  work  is  correct.     Try  other  numbers. 

50.  Monomial  x  Longer  Expressions.  The  product  of 
a  monomial  and  any  longer  expression  is  found  by  tlie 
same  process  ;   for  example  : 

4  a^.  [2^2  _  3  a6  +  5  52]  =  (4  a5)(  2  a^~)  +  ( 4  ahX  -  ^  «^) 

+  (4aJ)(5i2^) 

=  Sa%-V2(rb''+'20ab^. 
The  work  may  be  arranged  as  follows  : 

Multiplicand :  2  a^ -  3  ah  +  5  b"^ 

Multiplier :  4  ab 

Product :  8a% -\2aV)^  + 20  ab'i 

Check :        If  we  set  «  =  1  and  /^  =  1  as  in  §  40,  we  get 
4  ab[2  n^  -  3  ab  +  5  ft2]  =  4[o  _  ;3  +  5]  =  in  ; 
and  8  a'^b -  12  a^b^  +  20  ah»  =  8  -  12  +  20  =  10  (correct). 

Care  should  be  taken  not  to  overlook  negative  signs. 

78 


LOiNGKK   EXPRESSIONS  79 


EXERCISES  VII :  CHAPTER  IV 

IVrfonn  the  nmltiplications  indicated;  check  each  result  by 
substitution  of  numerical  values: 

2  a;  — 3  11.    x-  —  2xi/  +  y- 

5  xy ^ 

12.    6  x'y  —  3  xyz^ 

2  yz'' 


1. 

■^•  +  y 

a 

2. 

X  -  y 

a 

3. 

X  -  y 

—  a 

4. 

—  x-y 

—  a 

5. 

X  +  y 

—  2 

7. 

-  3  .1-  +  7 

2 

8. 

y-x-^ 

-3 

9. 

ax  —  by 

2  ax 

uO. 

vm-  —  m-n 

3  mn 

13.  a-yc-^abc" 
~alf 

14.  7.x-'-— 2;k  +  4 

-  3  .^•^ 

15.  ax  —  by  —  cz 

—  abxy 


16.   -3.i;*+6r''-4x--+5.T-l        17.  (a;-?/)H3(?/-2)-  +  (2-.x-)2 
-2.r^ (x-yf . 

18.  If  one  fifth  of  y  is  known,  how  may  y  be  found?  If 
^=-2,  y=?      lf^  =  x,y  =  ?     Uy^  =  x-io,y  =  ? 

19.  Plot  the  figure  for  the  equation  •l  =  x  —  l. 

51.  Division  by  Monomials.  Monomial  Factors.  To  di- 
vide an  expresaion  by  a  monomial  divide  each  term  by  the  mo- 
nomial and  connect  these  partial  results  by  the  proper  siyns. 

Thus,    (8  a%  -  12  ^2^2  +  20  ab^)  -  (4  ab)  = 

8a%    ,     -12a2A2        20  ^^3  o      l^nJ2 

1- -| =  2  a''  —  o  ab  +  o  f>^. 

4  ab  4  ab  4  ab 

The  truth  of  this  rule  is  evident  because  qundent  x  divisor  =  divi- 
dend :  thus,  the  example  just  solved  is  the  example  of  §  50  reversed. 

Care  should  be  taken  not  to  overlook  negative  signs.  In  checking 
divisions  by  substituting  numbers  for  letters,  be  careful  to  avoid  divi- 
sion by  zero  (p.  T.'i).     Tf  zero  occurs,  try  other  numbers. 


80  .MULTIPLICATION    AND    DIVISION  [Cn.  IV 

This  rule  is  chiefly  useful  in  finding  monomial  factors  of 
expressions.  Thus,  given  the  expression  ^6  a%  —  Vl  a^U^ -{- 
20  alr^^  we  can  see  by  inspection  that  4  ah  is  a  factor,  and 
we  write 

9,a%-l'2  a2^>2  +  20  ah^  =  4  abQl  a^  -  3  «6  +  5  h"^) . 

[Let  the  student  read  again  and  state  the  definition  of  factor,  §  8,p.8.] 

EXERCISES    VIII:  CHAPTER   IV 

Perform  the  following  divisions ;  check  each  result  by 
substitution  of  numerical  values : 


4. 


6. 


Factor  the  following  expressions  as  the  product  of  a  mono- 
mial factor  and  another  expression  : 

10.  ah  +  ac.  16.  2  a.ir'  —  6  a-x-  +  8  ah% 

11.  4:xy-\-6xz.  17.  15mV  — 20m-»''+25?>/V. 

12.  5 an/- 10 a^b\  18.  S ab\^ -  12 arh^c  +  10 aHrc. 

13.  6??iV  — 8mV.  19.  C^r\sH-  +  12r\^-f  — 18  r-s^t\ 

14.  _  9  ar"- 12  aV^.  20.  8ap\f +  2lhirq^-32p^f. 

15.  —  1 1  apq^  +  13  ap\  21.  —  10  x^yz  —  5  xyH  — 15  xyz'. 

22.  Factor  the  numerator  in  each  of  the  exercises  1-9. 

23.  Express  the  answer  in  each  of  the  exercises  in  Ex.  VII 
as  the  product  of  two  factors. 

24.  If  twice  a  number  is  known,  how  is  the  number  foiuid  ? 
If  2.r  =  12,  .r=?     If  2.T  =  -8,  .T=  ?     liax  =  a',x  =  ? 

25.  If  ax  =a-  —  2  ay,  x=?     If  ax  =  a-  +  ah  +  ay,  x  =  ? 


kx  —  3k               6  a^ 

-9  ah 

^      —  12  xy  —  16  ax  +  3  x 

k 

3a 

-4.x 

aV  -  5  a^'b''  -  a^b^ 

7. 

s^  —  as  —  bs  —  cs 

-a'b' 

3s 

2fx^  +  3gxy-7hxz 

8 

9aa^-12a'x'-27a^x^ 

2x 

—  3  ax' 

-\gt'-i9. 

q 

12  x-yh'  -  9  xy-z'  +15  xYz^ 

\9 

3  x'y-zr 

r,l-r,L']  LONGER   EXPRESSIONS  81 

26.  If  2  X  —  3  =  5,  a;  =  ?     If  ax  —  ab  =  ay,  x=? 

27.  12  .«  +  1()  //  =  48.  Keduce  to  simpler  form  and  plot  the 
figure  both  before  and  after  this  reduction. 

Simplify  : 

28.  (ax'"  +  bx")  -=-  X':  30.    {x"y"  —  2xy)^  xy. 

29.  (a.f"'//"  +  bi^y')  --  x'y\      31.    (o  .«'" "  - 15  a;'"-")  --  -  5  x-«-". 

52.  Product :  Two  Binomials.  In  tlie  product  of  two 
binomials,  for  example,  (a  +  6  )(c  +  c?),  we  may  write 
lii-«t  (a  +  ^)  (c  +  f?)  =  a  {c  +  d)  +  h  {c  +  c?) 

by  what  precedes,  if  we  regard  (c  +  </)  as  a  simple  quantity. 

Now,  a(^c  +  il)  =  an  +  ac?, 

and  h{c  +  (7)  =  he  +  ^>tZ. 

Hence,  (a  +  h)  (c  +  (7)  =  ac  +  ad  +  he  +  hd. 

Notice  that  this  is  true  for  numbers  of  various  kinds : 
(4  +  2)  (3  +  6)  =  (4  +  2)  •  9  =  4  •  9  +  2  .  9. 
But,  4  .  9  =  4(3  +  6)  =  4  .  3  +  4  .  6, 

2  .  9  =  2(3  +  6)  =  2  . 3  +  2  •  6. 
Hence,  (4  +  2)  (3  +  6)  =  4  •  3  +  4  •  6  +  2  •  3  +  2  •  6    (correct). 

As  an  example,  consider   (2  ab  +  3  6^(5  a  —  4  b). 

We  write  this  in  the  form  : 
Multiplicand :  2ab  +  Sb^ 

Multiplier :  .5  a  —  4  6 

Firsit  partial  product :  10  a%  +  15  ab'^ 

Second  partial  product :       —    8  ab'^  —  12  b^ 

Total  product :  10  a-b  +    7  ab'^  -  12  b^ 

Care  must  be  taken  not  to  ovei-look  negative  signs. 

In  viTiti ng  down  the  partial  products  we  write  in  one  line  the 
product  (2  ab  +  3  lfi)(o  a),  and  then  the  product  (2  ab  +  3  b-)(  -  4  h). 
Tlie  student  should  be  careful  to  write  similar  terms  in  these  products 
underneath  one  another,  so  that  they  shall  be  conveniently  placed  for 
addition . 


82  MULTIPLICATION   AND   DIVISION  [Cii.  IV 

53.  Product  of  any  Expressions.  The  product  of  longer 
expressions  may  be  found  by  a  similar  rule.  In  any 
case  we  multiply  each  term  of  one  of  the  expressions  by 
each  term  of  the  other  in  some  convenient  order  and  add 
the  resulting  partial  products. 

The  work  will  be  easier  if  the  terms  in  each  of  tlie 
given  expressions  are  arranged  first  in  a  definite  order ; 
thus,  if  some  letter  is  selected  that  occurs  in  most  of 
tlie  terms,  the  terms  may  he  arranged  so  that  the  exponents 
of  that  letter  increase  as  ive  go  toward  the  right  or  left. 

Thus,  to  multiply,         +  4  a;  -  3  x^  +  2  by  2  +  5  a;-  -  3  ar, 
we  rearrange  with  respect  to  the  letter  x  and  write : 
Multiplicand  :  2  +  4  x  —  3  x'^ 

Multiplier  :  2  —  3  x  +  5  x=^ 

First  partial  product :  4  +  8  x  —  6  x''^ 

Second  partial  product :  —  6  x  —  12  x'^  +    9  x^ 

Third  partial  product :  +  10  x'^  +  20  x^  -  15  x^ 

Product :  4  +  2  X  -    8  x2  +  29  x8  -  1.5  x* 

Check:  If  x  =  1,  multiplicand  =  3;  multiplier  =  4;  product  =  12 
(correct). 

An  arrangement  in  which  the  exponents  of  the  chosen  letter  in- 
crease as  we  go  toward  the  left  is  equally  good.  Tn  either  arrange- 
ment the  similar  terms  fall  under  one  another  if  we  shift  tlie  partial 
products  to  the  right  by  one  term  each,  provided  all  positive  integral 
powers  up  to  the  highest  are  present  in  both  factors.  If  any  terms 
we  should  naturally  expect  are  absent,  care  should  be  taken  to  put 
similar  terms  in  the  ]>artial  products  under  one  another. 

EXERCISES    IX  :  CHAPTER    IV 

Perform  the  following  miilti[)lications,  and  check  as  usual: 


1    a-  +  y 

a  +  b 

4.  ^~y 

'    —  a  — 

b 

7.  ^'  +  ^' 
a  —  b 

„      2x-3y 

8.  ^^ 
-x  +  5y 

9    a; +  2?/ 
2x-y 

10. 

11. 

12. 

a-  —  ab  +  b'- 
a  +  b 

2  ""-y 

a  +  b 
^-    a-b 

5.  "  +  ^ 

a  +  b 

6.  "-'' 
a  —  b 

(,2  _  ab  +  b' 
a  —  b 

a^  +  ab  +  \r 
a  +  b 

■,:i]  LON'CER  KXrKKSSIOXS  83 

13.  a'+ab  +  b'                        22.    2x'-3x'-x'  +  7X-S 
a-b  2  x-  -  1 

14.  .^-  +  3.i•-4                          23.    aV-2a-;K»-3a.c*  +  x' 
2x  +  '6  —  a-  +  ax-2x^ 


15.    -m^  +  L'//r-7  24.    2.jy^- 7.^-- 4.T  + 


9 


2  ?>i^  -  3 


X^  —  O  X  +  1 


16.  r^  +  5r-l                            25.    4?-''- 6  >"  +  ?•  + 1 
/•-  +  .3  r  +  4  -r  +  3r  +  2 

17.  xhf  —  ^xy^A:                    26.   3  7>i*— . 2  ?)i^+6m-+4  ?m— 3 
-3ayy  +  5       .  2m^-2m  +  3 

18.  l_4.'-  +  .j;2  27.    -2;i-^  +  3.r-5 
3  +  ;j.)--2.x-^  x'  +  x'  +  x  +  l 

19.  ;j.-2-3.iv/  +  2?/2                     28.    l-2.T-4-3a;^-4.r* 
3.T-2?/  1  -  .r-  +  2  -r 

20.  .1-  + 5-3.7;                           29.    .-j^^  +  2  —  3 .r- +  .r  —  4 .x* 
2  a;  -  6  +  •^r  a^  -  + 1  -  3  .r 

21.  j/  -  2  p(f  +  7  q^                     30.    r'  -  3  >"''<  +  2  J-^f ^  -  j-f'  +  t* 
p-  +  jx/  +  <f  4  r-  —  ^-  +  ?-^ 

Find  the  following  products  and  powers  : 

31.    {x-l){x-2){x-^).        32.    {x-a){x  +  a){x'+ce). 

33.  (a  -  5  b)  (a  +  Sb)  (a  -  b). 

34.  (a  +  bf.  36.    (a  +  by.  38.    (a-bf. 

35.  («  +  ?>/.  37.    (a -5)-.  39.    (a  -  6/. 

40.  (a  -t-  ft  +  c)  (a-  +  6-  +  c-  —  be  —  ca  —  ab). 

41.  (x2_4)(.r^  +  4).  42.    (.7;'  +  r)(''^'-r)- 

Show  that : 

43.  (.^■2  +  x>/  + .?/-)  (a;2  _  x>/  + .?/-)  =  x*  +  ■<'/■  +  ?A 

44.  (a-  +  '-'-) (•»'•'  +  f) -  («•«  +  ^!/)-  =  ("!/  -  'J-^')"^- 


84  MULTIPLICATION   AND  DIVISION  [(  ii.  IV 

54.  Division  of  Longer  Expressions.  Just  now  we  shall 
not  attempt  to  divide  one  expression  by  another  except  in 
a  few  easy  cases  in  which  the  quotient  is  known  to  be 
simple.  This  is  the  case  if  the  dividend  is  really  known  to 
be  the  product  of  the  divisor  times  some  simple  expression. 

Thus,  taking  the  example  worked  out  in  §  .53,  we  know  that  we  can 
divide  i  +  2 x  —  8x^  +  29 x^  —  lox*  hj 2  +  4:X  —  S x'\  and  we  know 
that  the  quotient  is  2  —  3x  +  5x'^  because  we  just  multiplied  the  latter 
two  expressions  and  found  their  product  to  be  the  first  expression  here 
mentioned.  If  the  quotient  were  unknown,  we  should  write  down  the 
following  scheme  (see  explanation  below)  :  „.  . 


Dividend :  4  +  2  x  -    8  x2  +  29  .r^  -  ISx" 


+  4  X  —  3  a;2 


2  —  '■]  X  +  D  x^ 


1st  Partial  Product:     i  +  8x  -    Qx^  (subtract)  Quotient 

\st  Remainder :  -  Qx  -    2x^  +  29 x^  -  15x* 

2d  Partial  Product:        -  6 x  -  12 x^  +    9 x^     (subtract) 

2d  Remainder :  10  x^  +  20  x^  -  1  •")  x* 

?yd  Partial  Product:  10.?;-  +  20 x^  -  lo-r^     (subtract) 

Final  Remainder :  0 

Check:  If  x  =  l,  dividend  =  12  ;  divisor  =  3  ;  quotient  =  4  (correct). 

The  final  remainder  being  zero,  the  division  is  said  to  be  exact. 

This  scheme  is  useful  in  rediscovering  the  partial  products  in  the 
work  in  the  example  of  §  53. 

The  explanation  is  as  follows  : 

(1)  The  first  term  of  the  dividend  divided  by  the  first  term  of  the 
divisor  gives  the  first  term  of  the  quotient.  Otherwise  the  work  of  §  53 
would  not  give  the  first  term  of  the  product  as  shown. 

(2)  This  first  term  of  the  quotient  X  the  divisor  is  the  first  partial  prod- 
uct in  §  53  ;  we  place  it  underneath  the  dividend. 

(3)  The  difference  between  the  dividend  and  the  first  partial  product 
must  be  all  the  rest  of  the  whole  product  in  §  53  ;  ice  therefore  subtract  them 
(result  called  "  1st  remainder"). 

The  next  steps  are  all  taken  for  similar  reasons  and  are  really  repe- 
titious of  the  above  steps  ;  they  are  : 

(4)  (First  term  oi  first  remainder)  -f  (fiist  term  of  divisor)  =  (second 
term  of  quotient) 

(5)  (Second  term  of  quotient)  x  (divisor)  =  (second  partial  product). 


-.4]  LOX(iER    KXrilKSSlOXS  85 

(6)  (First  remainder)  —  (second  partial  product)  =  (second  re- 
mainder). 

(7)  (First  term  of  second  remainder) -^  (first  term  of  divisor)  = 
(third  term  of  quotient,  last  in  this  example). 

(8)  (Third  term  of  quotient)  x  (divisor)  =  (third  partial  product). 

(9)  (Third  remainder)  —  (third  partial  product)  =  final  reinaiii<ti'r 
(=0  in  this  example). 

Steps  (1),  (2),  (:5)  are  simply  repeate<l  as  many  times  as  necessary. 
Thus,  the  steps  (4),  (5),  (6)  and  the  steps  (7),  (S),  (9)  are  the  same 
kind  of  steps. 

If  the  final  remainder  is  zero,  the  division  is  said  to  be 
exact;  this  always  happens  if  the  dividend  is  really  the 
(piotient  multiplied  by  another  simple  expression. 

If  the  remainder  is  not  zero,  the  division  cannot  be 
entirely  carried  out ;  in  that  case  what  is  left  over  is 
called  the  final  remainder,  or  simply  the  remainder.  This 
will  usually  happen  if  the  example  has  not  been  carefully 
selected  to  avoid  it. 

Thus,  if  we  try  to  divide  6  -  lOx  +  Sx^  -  Tx^  -  24x4  by  2  +  4  x 
—  3  x'\  the  work  is  as  follows  : 

Dividend ;  6  - 10  x  +   3  x^ -      7  x^ -  24  x^  2+    4x-    3x2      Divisor 
1st  p.p.       6  +  12x-   9x^      (subtract)      3-11  x  +  28x^     Quotient 
1st  rem.         -22  x  + 12x2-     7x3-24x'' 
2d  p.  p.         -22x-44xH   33x8     (subtract) 
2d  rem.  56x2-   40x3-24x* 

Zdp.p.  56x2+H2x8-84x-'     (subtract) 

— 152  x^  +  (JO  X*     Final  Remainder 

In  this  case  the  division  is  "  not  exact  "  since  the  final  remainder  is 
not  zei-o.  Just  as  in  arithmetic,  we  may  still  express  the  result;  tliiis 
31  ^  7  gives  quotient  4  and  remainder  3  ;  we  say 

,.  . ,      ,        ...  ,■     ,    ,   remainder 

dividend  -^  divisor  =  intotient  +  • ^-^ , 

divisor 

i.e.  31  -f-  7  =  4+  f.     Similarly,  here  we  say 
(6  -  10  X  +  3  x2  -  7  x3  -  24  x"*)  -  (2  +  4  x  -  3  x^) 

(dividend)  (divisor)  (remainder) 

—  l.^'^x^  4-  60  X* 
=  (3  -  1 1  X  +  28  x2)  +       1^-^   +b"|  . 

(quotient)  /]•   •      N 

(divisor) 


86  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

It  is  absolutely  necessary  for  all  this  work  that  both  divi- 
dend and  divisor  be  carefully  arranyed  in  the  same  order 
(eitheT  exponents  increasing  to  the  left  or  to  the  right) 
with  reyard  to  the  same  letter. 

In  the  examples  just  worked  the  exponents  increase  to  thn 
riyht ;  we  may  work  the  same  examples  equally  well  witb. 
exponents  increasiny  to  the  left. 

The  work  for  the  example  of  p.  83  follows : 

-  3  a:2  -f  4  X  +  2      Divisor 


5x^  —  3  a,'  +  2      Quutieni 


-3,y2+    4a:+    2     Dirisor 
8a-^  +  13x  +  ^^^      Quolieni 


Dividend :  -  15 a;*  f  29 x^  -    Sx^  +  2x  +4 
-  15x''  +  20.r3  +  10x2 

9  x3  -  18  x2  +  2  X  +  4 
9x3-  12x'^-6x 

-  6  X-  +  8  X  +  4 

-  6  x'^  +  8  X  +  4 

U     Final  Remainder 
Notice  the  work  is  really  the  same  as  before ;  the  answers  also  are 
in  this  case  the  same,  but  with  tlie  terms  in  reverse  order. 

But  if  the  division  is  not  "exact,"  the  work  and  the 
answers  change  when  we  change  the  arrangement. 

Thus,  in  the  first  example  on  p.  85 : 
Dividend:  -24x4-   7^.3+    3^:2_i0x+0 
-24x4+32x3  +  16x''^ 

-39x3- 13x2- 10  X +  6 
-39x3  +  52x24-26x 

-65x2-36x  +  6 
-65x2+g|Ox+  H° 

—  ^1^  X  —  '^\'^     Final  Remainder 

This  work  is  totally  different  from  the  work  done  before  on  the 
same  example.  It  is  for  this  reason  that  very  few  examples  of  this 
kind  are  given  below.  However,  the  student  need  not  be  surprised; 
f(n-  example,  31 -j- 7  =  not  only  4+f;  but  also  3  +  y,  and  5  — f 
Moreover,  in  the  form  on  p.  85,  the  division  may  be  carried  to  further 

terms,  if  desired,  just  as  31  -^  7  =  4.42  -[■'- — ,  in  arithmetic. 

If  several  letters  occur  in  a  problem,  one  of  them  —  usually  thf> 
most  prominent  one  —  is  used  for  the  purpose  of  arrangement.  It 
another  letter  is  used  for  arranging  the  expressions,  the  work  may  be 
wholly  different,  but  the  results  will  always  be  precisely  the  same  if 
the  division  is  exact.     Thus : 


54]  LONGER   EXPRESSIONS  87 

Dividend : 

2  X*  -  17  x^y  +  31  x^  -  23  xi/^  +  12  y*i2x-Sij Divisor 

2x*  —  3  x'^y  I  x-^  —  7  x'^y  +  5  xy^  —  iy^   Quotient 

-  n  x^y  +  31  xY  -  23  xy^  +I2y* 

-  14  x^y  +  21  xY 

10a:V-23ay+  12  y* 
10  x'Y  —  1 5  x/y^ 

-Sxy^+V2y* 
-  8  X//3  +  12  y^ 

0     Final  Remainder 
Check:   If  x  =  1  and  y  =  1,  dividend  =  5;  divisor  =  —  1;  quotient 
=  —  5  (correct). 

The  student  will  be  able  to  work  the  exercises  that  follow,  after 
carefully  studying  these  examples  and  trying  to  do  them  over  him- 
self without  looking  at  the  book.  The  examples  in  which  the  divi- 
sion is  "  not  exact"  are  marked  so  that  the  student  will  see  them  in 
advance.  Some  of  the  examples  are  not  arranged  as  they  stand ;  the 
student  must  in  all  cases  see  to  it  that  both  dividend  and  divisor  are  prop- 
erly an-anged  before  he  tries  to  do  the  exercises. 

After  some  experience  the  student  will  be  able  to  omit  parts  of  the 
work  shown  above ;  but  it  is  best  not  to  do  this  too  soon.  Later 
(Appendix,  §§  1-4)  we  shall  show  how  to  reduce  the  labor  very  much. 

EXERCISES   X:    CHAPTER   IV 

Perform  the  following  divisions.  In  the  exercises  marked  * 
the  division  is  not  exact,  and  the  given  arrangement  of  terms 
should  be  used;  the  others  should  be  suitably  rearranged,  if 
necessary. 

1.  (ct'-h-)^{a-h).  9.  (x--l)H-(a;-l). 

2.  (a;--3a-+2)H-(x'-l).  10.  (m'^-«-)--(m-f »). 

3.  (3x-2+5i«-2)H-(a;  +  2).  .11.  (o^ -y')^{x- y). 

4.  (a2-12a&-13  62)H-(o-13fe).         12.  (/•3-(-l)^(v  +  l). 

5.  {ah-ac-hd  +  cd)^{a-d).         13.   (m''  +  8)--(m -f-2). 

6.  (^•«-8P-f-16)^(Ar^-4).  14.  {x^-l)^{x-l). 

7.  {ie-?,a-h  +  ^ah--ly')^(ci-h).  15.  (m'' -//*)-- (m  +  m)- 

8.  (x-^-f  4a-^//-  +  4?/^)-(.r  +  2/).        16.   {ic' -  y')-^{x -y). 


88  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

17.  (A'  +  3  A-B  +  3^7?-'  +  B') -=- (.4-  +  2AB  +  B'). 

18.  {2-x-  ix-  +  17  x^ _  12 X*) -- (2  +  3  X -  4 x'). 

19.  (6  a;^  -  2  x'  -  8  ar  +  4  a;  -  8)  --  (3  .^•2  -x-\-2). 
CO.  (()  -  2  >M  -  8  /u-  +  4  m^  -  8  m')  --  (2  -  4  m-). 

21.  (8  X'  +  1 0  a;"  +  r>  XT  + 15  a-'  -  12  .<•  +  2)  h-  (2  a;^  +  3  x-  -  1). 

22.  (4  /'  +  7-'  +  2  /•*  -  4  j-2  -  r  -  2)  -  (4  /-  -  3  r-  +  r  -  2). 

23.  (2-17/-+4G'/-'-43y-  +  20r-'Vl8  r^+8  /•«)-(l-4r+2r-). 
•     24.  (24  ?/*'  -  28  /  +  14  //"  -  10  jf  +  r>  ?/2  _  1)  ^  (4  2/2  _  2  ^/  _  i). 

25.  (4  +  42>  +^j2  _  3 ^^3  _  2^j4  _|_  ^,-'  _|_y!)  H- (4  -  3i>-  +  ^j^). 

26.  (2  .^-s  +  a;^y + 4  a.-^"  -  2  x^  -  xf  -  4  ?/«)  -  (.i- ■ + xhj + x-^'-'  +  f) . 

27.  (a;^  +  7  ar^  +  14  a;  -  15  -  11  a;^  -  2  .^•*)^(-  a;  +  3  +  x"). 

*  28.    (x-^  -  x-  -  7  X--'  +  28  X-  -  11 )  -  (x--'  -  4  X-  +  7). 

29.  (x-2  —  a  +  b  •  X-  +  a?>)  ^  (.«  -  a).    *  33.  x''  -^{x^  —  x-2). 

*30.  (x''-2x-^-10x-)h-(x-2-3x  +  1).    34.      (y/«-l)--(2/-'_l). 

31.  (a*  +  i*  +  cr6-)-^(a2+ ?y--a/>).     35.  (m^'—n^)-^{7n'^  —  n^). 

32.  (m^  +  32H^)-(/>i+2n).  *  36.  (F- 4)-(^'^- 1). 

37.  {R^  -  6  R'  +  11  Ji  ~  6)^(R-  -  5  7^  +  G). 

38.  (x'^'  +  a;  +  5  -  2  X'''  -  7  x'2  +  2  x" )  H-  (2  x""  -  5  -  x-  +  ar^). 

*  39.    (v-^  -  4  7-*  +  5  7-3  -8  7-2  +  8  7-  -  C)--(7-^-  37-2  +  2  r-  7). 

55.  Polynomials.  The  expressions  in  the  exercises  given 
in  this  chapter  are  all  very  simple.  They  are  in  fact  poly- 
nomials.     (See  §  9,  p.  9.) 

The  most  general  polynomial  in  the  single  letter  x  is 

a  +  bx+  cx^  +  dj?  +  •  •  •  +  ?a;" 

where  a,  5,  <?,  •••,  Z  are  constant  numbers,  possibly  zero,  and 
where  7i  is  some  positive  integer. 


M-5o]  LONGER    EXPRESSIONS  89 

The  most  general  polynomial    in   two  letters,  z,    ?/,    is 

a  +  hx  +  cy  +  dx^ ->r  ery  ^fy"^  +   ... 

(a  finite  number  of  terms),  where  a,  b,  c,  •••  are  constant 
numbers,  possibly  zero. 

Each  term  of  a  polynomial  contains  each  letter,  if  at  all, 
only  as  a  factor  which  is  a  simple  power.  If  a  letter  oc- 
curs under  a  radical  sign,  or  in  the  denominator  of  a  frac- 
tion, the  whole  expression  is  not  a  polynomial  in  that  letter.* 

Thus,  3  —  22;-f5a;^isa  polynomial  in  x,  4  my^  +  6  m^y  — 
7  m'^  is  a  polynomial  m  y  and  m.  5  x^y  +  3  x'^y  is  a  poly- 
nomial in  X,  but  is  not  a  polynomial  in  y  (because  \/y  is 

not  a  simple  power).  The  expression  2a;2  — 3-f--  is  a 
polynomial  in  x,  but  is  7iot  a  polynomial  in  y. 

All  the  expressions  in  this  chapter  are  polynomials  in 
all  the  letters  in  them,  except  some  few  exercises  that 
contain  fractions  or  contain  letters  as  exponents.  There 
will  later  be  many  examples  that  are  not  polynomials,  in 
the  Chapters  on  Radicals  (Chapters  VII,  XI)  and  in  the 
Chapter  on  Fractions  (Chapter  V).  The  distinction  is  of 
especial  importance  in  Factoring.     (See  p.  91.) 

REVIEW  EXERCISES  XI:  CHAPTER  IV 

Perform  the  indicated  operations ;  check  as  usual : 

1.  (a-  — a +  !)(«-  + a -fl).  3.    [x- —  (a-{-b)x+ab^(x —  c). 

2.  (v-5)('y-t-3)(«-2).  4.    (a'  +  b^'-aby. 

5.  {m^-  2  »i-  +  m  -  3)  {m^  -  3  m  -  2). 

6.  (2x-  3  y/)(3  y-o  z)(d  z-2x). 

7.  (3  a.-2  -  4  X  +  o)(2  .r*  -  3  .-r^  -  4  a;  -  2)(1  -  2  .t  -  3  x^). 

*  Although  many  elementary  text-books  define  this  word  no  as  to  in- 
clude any  kind  of  terms  whatever,  the  author  of  tliis  book  can  find  no 
standard  authority  for  any  other  detinilion  than  that  here  given.  See,  e.g., 
Bocher,  Introduction  to  Higher  Algebra,  pp.  1-4. 


90 


MULTIPLICATION    AND   DIVISION 


8.  (ar'  +  xy  +  if)  (x"  -  xy  +  y^)  {x'  -  .r  y-  +  y'). 

9.  (j-2  -f  2  r.s  -\-A.e){r^-l  f) (/•-  -  2  r.s  +  4  ^2^,. 

10.  (.e'  -  5  x'y  -  6  ar^?/2  +  .>%^  -  2  .tv/'  +  f)  (3  a.-  -  2  ^^j. 

11.  (2  a;  -  X-  + 1)  (ar^  +  2x-x'  +  4)  (.^•  +  3). 

12.  (i97  +  5' + p-)  (9'  +p'  -  ;^?) iSf  -  P')- 

13.  (8  Imn  +  2  f  -  4  Z-//i  +  2  In)  (3  ?//,  +  ^-  -  m). 

14.  (aa;  —  hy)  (by  —  cz)  (cz  —  a.i;) . 

15.  (aV  —  a^a;?/  +  &-?/^)  (bY  —  bcyz  +  c-2-)(c-2-  -  cazx  +  a^x^). 


16. 


y  +  a;  ar  +  a;  —  1 


17. 


,.6  _  ^6 


* 


19. 


Show  that ; 


x^+2x*-^ 

ar  +  1 


20. 


21. 


ar'  4-  y^  —  6  .t?/  +  8 
x-—  3;?/  +  ?/^— 2x— 2?/4-4 

a;3_9a.2^23a;-15 
a.-^  —  6  a;  +  5 


22. 


X"  -  y 


(9  m^  +  3  m7i  +  7^^)(9  m^  -  3  m»  +  w-)(9  r>r  -  u-)  _  ^ 
(27  7/1''  -  71^) (27  ?;r^  +  y*'^) 


24. 


(g^  -\-b'-\-c-) (xr  +  y-  +  z-)  -  (aa;  +  %  +  czf  ^  -j^^ 
(ay  —  bx)--\-{bz  —  07/)-+ (ca:  —  a^)- 


25.  Divide  a.-^  -  3  x^  +  5  a;  —  7  by  x-a,  keeping  coefficients 
of  like  powers  of  x  together.  Arrange  your  remainder  as  a 
polynomial  in  a  with  exponents  decreasing  to  the  right.  What 
do  you  note  about  the  remainder  ? 

26.  What  would  be  the  remainder  ondividiugar''— 3  ^^+5  x—7 
by.x--l?  bya;-2?  by  .^•  + 1  =  a^- (- 1)? 

27.  Divide  a:^  —  ar^  +  3  a;  —  ?)i  by  a;  - 1.  What  is  the  remain- 
der?   How  may  m  be  chosen  so  that  the  division  may  be  exact  ? 


28. 


a.-2  +  (a;  +  l) 


29. 


l-3(l-a;)+2(l-a;)^ 
l-(l-.'r) 


PART   III.     SPECIAL   MULTIPLICATIONS;   FACTORS; 
TYPE    FORMS 

56.  Introduction.  There  are  some  examples  that  occur 
so  often  in  applications  of  algebra  tliat  it  is  desirable  to 
commit  them  to  memory. 

As  an  example  of  what  is  to  be  done  liere,  consider  {x  +  yy^,  which 
means  (x  +  y){x  +  ij).     Actually  multiplying,  we  get, 


x^ry 
X  +  y 

x^  +  xy 

xy  +  y^ 

x2  +  2  xy  +  .y2 

Then  (x  ■\-  y^  =  x'^  -{■  2  xy  +  y'^.  This  holds  whatever  x  and  y  may 
mean.  Thus  (4  +  tj)^  =  4'^  +  2 (4  •  6)  +  6^  =  100.  Again  (/«  +  n)-  = 
11  fi  +  2  mn  +  n'^. 

To  use  this  (or  any  other  forms  in  this  chapter),  we  try 
to  see  whether  a  given  example  can  be  made  to  look  exactly 
like  the  known  example  by  pairing  off  the  parts. 

Thus,  (3a  +  2hy^  is  exactly  like  {x  +  y)-  if  x  —  'ia  and  y  =  2b. 
Hence,  since  (x  +  y)'^  =  x'^  ■{■  2  xy  +  y-, 

(3  a  +  2  by  =  (3  ay  +  2  (3  a)  (2  h)  +  (2  i)2  =  9  a-^  +  12  ah  +  4  ^2. 

C/iecit;  If  a  =  1  and  6  =  1,  (3  +  2)2  =  25  =:  9  +  12  +  4  (correct). 

We  try  to  remember  the  answer  for  anotlier  reason  ;  for 
(a;2  j^2xy  -{-  3/^)  -^(x  +  y')  =  x  +  y;  if  this  example  in  divi- 
sion is  given,  it  is  convenient  to  know  the  answer. 

Finally,  we  say  the  factors  of  x^  +  '2  xy  +  y'^  are  {x  +  y) 
and  (x  +  /y),  for  the  product  of  these  factors  is  the  given 
expression.     (See  p.  9.) 

The  expressions,  x'^  ^'ixy  -{-  y^,  x-\-y,  x  +  y  are  all  poly- 
nomials in  the  letters  x  and  y.  (See  p.  88.)  In  general, 
in  this  chapter  and  throughout  the  book,  we  shall  seek 
only  for  polynomial  factors  of  polynomials. 

91 


92  MULTIPLICATION   AND   DIVISION  [Cii.  IV 

Tliis  is  similar  to  the  custoiu  in  aiithiiietio,  whei'e  we  usually  seek 
only  for  integral  factors  of  iiilegers ;  thus,  we  say  the  factors  of  10  are 
2  and  5 ;  we  do  not  say  the  factors  of  10  are  |  and  6,  for  example, 
though  I  X  6  =  10. 

Ex.  1.    Find  the  factors  of  4  nv  +  20  mn  +  25  ri^. 

Given -4  m"'^  +  20  m«  +  25  m'^,  we  may  ix)ssibly  notice  that  this  is  the 
same  as  (2  in)'^  +  2  (2  m) (o  n)  +  (5 n)'^.  Comparing  witli  x^  -\-  2x  ■  y 
+  y'^,  we  see  that 

4  m^  +  20  mn  +  2.")  n^  -  (2  m  +  5  n^. 

The  factors  of  4  m^  +  20  mn  +  5  ifi  are  therefore  (2  m  +  5  n)  and 
(2  m  +  .5  n)  ;  in  other  words,  4  «i^  +  20  mn  +  25  n^  is  /Ae  square  of 
(2  m  +  5  n).     Check  by  actually  multiplying  (2  m  +  5  n)  by  (2  jn  +  5  n). 

Ex.  2.    (21)2  ^  (^20  + 1)2  =  202+  2  (20  . 1)  +  V  =  441. 
This  will  be  found  an  easy  way  to  square  certain  numbers. 

Ex.  3.   {a  +  h  +  c)-  =  [(Vt  +  h)  +  c]2=  (a  +  hf^  2(a  +  6)  .  c  +  c^ 
=  a^  +  h-^  +  c--\-2  ab  +  2  ac  +  2  6c. 

EXERCISES   XII:    CHAPTER   IV 

Perform  the  following  multiplications  and  divisions ;  check 
by  substitution  of  special  values : 

1.  (2A:  +  3c)2.      3.  (i/  +  iy-.         5.  (.»-^  +  3)2.        7.  (lx  +  \z'). 

2.  (.^•  +  2)2.  4.  (m  +  3  2;)2.       6.   (.^•^  +  2  .r)-.      8.   (.-»•  + 7.3)2. 
9.  (a +  2  6+5)2.  i3_  (a^.' +  24  a;+ 144) -(.«)+ 12). 

10.  (2  7/i  + // +i9)2.  14.  (i22+i6i^  +  64)-(/?  +  8). 

11.  U-^^  +  3  a- +  2)2.  15.   0j2  +  3;>ry  +  2ir/)H-0j  +  ff/). 

12.  ((m;  +  67/ +  C2)2.  16.  (9a;*  +  12.r'  +  4iK2)--(3a;,+  2ic). 
17.  (23r=(20+3)2;  112;  322;  (^^o^^^i^.  (7.2)2=  (7 +  .2)2. 
la.  (a  —  6)-.       [Work  this  by  writing  (a  —  h)  =  a  +  (—  /').] 

19.  Of  what  expression  do  you  suspect  9  b-  +  ('>b-\-l  to  be 
the  square  ?     What  are  the  factors  of  9  ^2  +  G  6  +  1  ? 

20.  Of  what  expression  do  you  suspect  .^•2  +  14i^•  +  40  to  be 
the  square  ?     What  is  the  square  of  that  expression  ? 


5(i-;,bJ  FACTORS;   TYPE   FORMS  93 

Resolve  into  factors : 

21.  a'  +  2ci?h  +  h\  25.  2>'  + 12  irz+ZQz\ 

22.  //- +  50 // +  625.  26.  DOl  =  1)00  +  60  + 1 ;  441 ;  484. 

23.  r  +  2  ?•*•  4-  *'-.  27.  a-2:^  +  26  ax_j/z  + 169  ary-. 

24.  Z2  +  82  +  16.  28.  (o-4-2a?;  +  6-)  +  6(a  +  ^}c  +  9o-. 

57.  Square  of  Sum.     The  result  of  §  56  is  : 

The  square  of  the  sum  of  two  lerins  equals  the  square  of  the  first  term, 
plus  twice  the  product  of  the  two  terms,  jdus  the  square  of  the  second  term. 

58.  Square  of  Difference.      Likewise 

Actually  multiply  (.r  -  i/)(x  -  >/)  to  get  this  result. 

The  square  of  the  difference  of  two  terms  equids  the  square  of  the  first 
term,  minus  twice  the  product  of  the  two  terms,  plus  the  square  of  the  second 
term. 

Ex.  1 .    (4  A;  -  3  d)-= (4  Af  -  2  (4  h)  (3  s)  +  (3  sf 
=  16  A-  -  24  A-s  +  9  S-. 

Ex.  2.     (18)2  ^  (20  _  2)2  =  (20)2  ^  2  (20)  (2)  +  2^ 
=  400  -  80  +  4  =  324. 

This  will  he  found  an  easy  way  to  square  certain  numbers. 

Ex.  3.    (m-  —  2  m.n  +  n-)  -e- {m  —  n)  =  m  —  n. 

This  results  from  the  formula  above. 

Ex.4.    (a  +  Z>-c)-  =  [(a4-'>)-c]' 

==  cr-  -^2ab  +  h'-2  ac  -  2  be  +  c^ 
=  a-  +  b-  +  c'  +  2  ab  —  2ac  —  2  be. 
Ex.  5.    Find  the  factors  of  4  m^  —  20  mn  +25  ir. 
As   in  Ex.  1,  p.  02,  we   notice  that  the   given  expression  may  be 
written  (2  m)-—  2(2  m)(')  n)  +  (5  n}^.     Comparing  with  II,  we  see  that 
4  /«2  —  20  nin  +  25  n^=z  (2  m  —  6  n)-. 
Hence,  the  factors  of  4  m~  —  20  mn  +  25  n^  are  2  m  —  5  n  and  2  m  —  5  n. 
Check  by  actual  multiplication. 


94  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

EXERCISES   XIII:    CHAPTER    IV 

Perform  the  following  multiplications  and  divisions ;    check 
as  usual : 

1.  (5r-2s)\     3.   (-a  +  bf.      5.    (v-2  tuf.    7.    (x-27j+2y. 

2.  (9&-2)-.       4.    (2v-13uy.   6.    (x'^-7ixy.     8.    {x-y-zf. 
9.    (//-10  6  +  25)^(6-5).      14.    l{x  +  y)  +  {x-y)J. 

10.  (16-8A;  +  fc2)--(4-A-).       15.  l{x+ y)-{x-y)Y. 

11.  (64ic2-16.r?/  +  ?/-)-T-(8.K-?/).  16.  (a-?>-c  +  d)2. 

12.  (m2-9m  +  20|)--(m-4^).   17.  (o  +  6)2  =  [«-(- 6) 7. 

13.  (a;''-8a-2+16)-(x2-4).     18.  (19)-  =  (20-l)2;  (29)'';(9)^ 

Resolve  into  factors : 

19.  A;2-6A:  +  9.  24.  4e2_96e  +  576. 

20.  2''-12z'  +  36.  25.   64pY-80i?9  +  25. 

21.  r-- 14^  +  49.  26.   16.T«-40a.r^  +  25a2. 

22.  ^)- +  9 «- —  6  7j«.  27.    9  a^a;2  —  42  atey  +  49  &y. 

23.  ?>i2_  24??u(+144n^  28.    {m^—2mn  +  n^—2{m  —  n)p-^p''. 

59.    Product  of  the  Sum  and  Difference.     Likewise, 
III.    (x+/)(jr-/)=jr2-y2. 

Actually  multiply  (x  +  y)  by  (a:  —  ?/)  to  get  this  result. 

Tlie  product  of  the  sum  otid  the  difference  of  the  same  two  terms  eqiials 
the  sijuare  of  the  Jirst  term,  minus  the  square  of  the  second  term. 

Ex.  1.    (3a  +  2ft)(3a-26)  =  (3a)--(26)2  =  9a--4ftl 

Ex.  2.    (a  +  /j  +  c) (o  +  6  -  c)  =  \_{a  +  h)  +  c]  [(a  +h)-c] 

=  (a  +  6)2  -  c2=  a?-\-2ah  +  b^-c". 
Check  by  actually  multiplying. 

Ex.  3.    Find  the  factors  of  4  a^  -  b'i     We  write 

A  a'-  b-  =  {2c(f  -b''  =  (2a  +  b){2a-b). 

Hence,  2  a  -\-  b  and  2a  —  b  are  the  factors. 


58-60]  FACTUUS;  TYPE    FORMS  95 

EXERCISES   XIV:    CHAPTER   IV 

Perform  the  following  multiplications  and  divisions;  check 
as  usual : 

1.  (4:a  +  h)(ia-b).  8.  (a- b  -  c)(a  +  b  +  c). 

2.  {2x  +  3y)(2x-3y).  9.  (a  +  '2b-Sc){a  +  2b +  3f). 

3.  (.)r^+5)(.i;2-5).  10.  (z'' -Ur')^(z-ir). 

4.  (.»r'  +  5)(-.r^  +  5).  11.  (a* -b'')^(a' +  b^). 

5.  (m  —  10  kz)  (m  +  10  kz).        12.  (7n'ii*  —  m*)t^)  -^  (nm^  +  7nhi). 

6.  (rt  -  &  -  c)  {a  +  b-  c).         13.  (25  ?;'  -  49  V2^^(5  n —1  m). 

7.  {a-b-c){a-b  +  c).        14.  (64/(/-- 2r))H-(8i>g -(-5). 
15.  99 -f- 11  =(100-1) -(10  +  1).  16.   99 --9.     17.   624-26. 
18.  24.  26  =  (25-1)  (25  +  1).  19.    15-17.     20.    35-37. 

21.  (a^  +  2ah  +  b-  -  16  c')  -  (a  +  &  -  4  c) . 

22.  (a^  +  6  ab  +  9^-  -  4  x-^  +  4  xij  — ;/-)  -^{a +  3b -2x  +  y). 

Resolve  into  factors: 

23.  100?-2-169sl  27.  a='-2rt6  +  6--9cl 

24.  4.4^  —  9  0-.  28.  x'^—ar. 

25.  .r*-64x-^.  29.  y*-b-  +  2bc-c^. 

26.  a- —  2  a.K  +  .1-2  —  4 /.  30.  16  — a;y. 

31.  z*  +  z-  +  l  =  (z*  +  2  2^  +  1)  -  Z-. 

32.  a-  +  2  o/>  +  />-  -  c-  +  6  cy  —  9  //-. 

33.  a-  -2ab  +  ?y'-  4  Z^  + 12  /«i  -  9  m\ 

60.    Likewise, 

IV.    (r  +  a)(x+6)=jr2+ (o+6)r  +  a6. 

Actvially  multiply  (a:;  +  a)  b}^  (a;  +  6)  to  get  this  result. 

The  product  of  two  binomials  whose  first  teiin  is  the  same  equals  the 
square  of  the  common  term,  plus  the  sum  of  the  second  terms  times  the  com' 
mon  term,  plus  the  product  of  the  second  terms. 


90  MULTIPLICATION   AND   DIVISION  [Cn.  IV 

Ex.1.    (a;  +  2)(.T  +  4)  =  a^  +  6a;  +  8. 

Ex.  2.    {x  -  3)(x  -f  5)  =  [aj  +  (-  3)]  [■'•  +  r,] 

=  .^-^'  +  [(-3)  +  o]..+  (-3)(5) 

=  £■  -\-'ix  — 15. 

Notice  that  the  rule  really  applies  to  differences  as  well  as  to  sums, 
as  iu  example  2.  We  need  only  express  the  difference  a;  —  3  as  the 
sum  of  X  and  —  3. 

On  the  same  principle  (x  —  yY  can  be  worked  out  by  the  rule  for 
(x  +  y)2.     Thus, 

(  X  -  ?/)2  =  [.r  +  (-  y)-f  =  x'^  +  2x(-y)  +  (-  yY  =:  x^  -  2  xy  +  y\ 
which  is  the  same  as  the  result  in  §  52.     (See  Ex.  18,  List  XII.) 

This  rule  is  used  chiefly  to  find  the  factors  of  (/iven  ex- 
pressions like  the  products  above. 

Ex.  3.    Find  the  factors  of  a;'+  6  x  +  8. 

We  write  a;2  +  6  x  +  8  =  (x  +  ?)(x  +  ?). 

Now  the  last  terms  have  a  product  8.  Hence,  we  try  such  combina- 
tions as  1  and  8,  and  2  and  4,  etc.  The  pair  1  and  8  is  not  cori-ect,  for 
(x  +  1 )  (x  +  8)  =  x^  +  9  X  +  8,  which  is  not  the  given  ex[>ression.  In 
fact,  it  is  clear  that  the  sum  of  the  numbers  of  the  cori-ect  pair  must 
be  6.  We  want,  then,  a  pair  of  numbers  whose /)?Wwc<  is  8  and  whose 
Slim  is  6.  If  we  try  a  few  pairs,  we  shall  probably  try  2  and  4  quite 
soon;  this  pair  is  correct,  for  2  x  4  =  8  and  2  +  4  =  6.  Checking  the 
answer  by  multiplication,  we  find  (x4-2)(x  +  4)  =  x'-^  +  G  x  +  8.  Hence, 
the  required  factors  are  x  +  2  and  x  +  4. 

If,  after  trying  all  pairs  whose  product  is  8,  we  find  no  pair  that 
is  correct,  we  must  give  up  the  problem  just  now  ;  later  we  shall  solve 
such  problems  by  a  different  method. 

Ex.  4.    Consider  cc^  +  2x  —  15,  the  resvdt  of  example  2. 

We  must  choose  a  pair  of  numbers  whose  product  is  —  IT).  Sucli 
pairs  are  —  1  and  +15,  +1  and  —  15,  +  3  and  —  5,-3  and  +  5.  But 
the  sum  of  the  pair  nuust  be  2;  hence  the  last  pair  is  the  correct  one 
since  —  3  +  5  =  2. 

Check :  (x  —  3)(x  +  5)  =  x-  +  2  x  —  15.  The  required  factors  are 
therefore  (x  —  3)  and  (x  +  5). 

Ex.  5.    Similarly,  1  +  2  cc  - 1 5  a-^  =  (1  _  3  a')(l  +  5  x), 
and  x^  +  2  xy  — 15  if  ={x  —  3  _?/)  (.c  -|-  5  y). 

The  letters  used  should  not  confuse  the  studout. 


0(1]  FACTORS;    TYTE    FORMS  97 

EXERCISES   XV:    CHAPTER   IV 

Perform  the  following  multiplications  : 

1.  (ry+2)((/-3).  11.    (pq-S)(pq  +  l). 

2.  (z  +  l)(z  +  4:).  12.    (t-Sa){t-9a). 

3.  (m-2n){m.  +  yi).  13.    (xf-2)(xy'- +  10). 

4.  (^r-s)(r-5s).  14.    (.s  +  ^  _  l)(.s  +  ^  -  3). 

5.  (.^•-3)(.^•  +  5).  15.    (a;?/2  +  2  0(a;?/2;  +  4  0- 

6.  (a  +  6)(a  +  2).  16.    (1  -  2  c'')(l  +  3  c^). 

7.  (l-s)(H-3s).  17.    (a--i)(.'r-f). 

8.  (;  +  2m)(?  +  7m).  18.    (6  -  f  a)(6  +  f  a). 

9.  (i'2-2)(v2-9).  19.    (2z-9)(2  2  +  8). 

10.  (a-3  +  3)(af'-4).  20.    (6p9  -  2  rs)(6iJ9-3  rs). 

Resolve  into  factors: 

21.  2- +  a2 -()((-.        24.    t'-2t-3.  27.    a-2-6a;  +  5. 

22.  a-  +  6  a  +  8,  25.   a--  —  5  .c  +  6.  28.    .r  +  6  .«  +  5. 

23.  a-  —  6  a  +  8.  26.    .r-  +  5  .i-  +  6.  29.    1  -  3  2;  -  18  2I 

30.  ?/i- -  2  m?t  —  15  n-.  33.    y- —  14:  xy  +  24  x\ 

31.  ?•-  — 12r6«  +  35s2.  34.    ?«^  +  5  m^n  —  24  7>i  V. 

32.  1  —  12  AS  +  35  7V.  35.    ?- — 15  ^«<  +  56  <r. 

36.  4u'-12u  +  5  =  (2  w)2  -  6  (2  u)  +  5. 

37.  9a^-|-6cK-8.  43.   x^tf -  15 xyz - 34 z\ 

38.  25  +  15  a;  —  4  x-2.  44.    x- —{z  +  2)x  +  2z. 

39.  4-8p9-21pV-  45.   ccy22_(-3_^)a.y2_3^ 

40.  ciH-  — 11  a^  —  26.  46.    a-  —  (x  —  y)a  —  xy. 

41.  c^d?-13c-de-  +  40e*.  VI.  f  +  1  fr -30i^. 

42.  lQx'-32xz  +  15z-.  48.    25  - 10  g- 48  92. 

49.  x'-2  xy  +  ?/2  _  9  (a-  -  ^)  +  20  =  (a-  -  ^)2  -  9  (x  -  y)  +  20. 

50.  ^y^  +  Q  pq  +  <d  q^  +  5{p  +3q)-14 

=  0'  +  3fy)-  +  50j  +  39)-14. 


98  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

61.   FormV:   ax^  +  bx  +  c.   The  product  (2  a; +3)  (4  a: +  5) 

=  8  2;^  +  22  a;  +  15,  as  is  seen  by  multiplying  (2  a:  +  3)  and 
(4  a; +  5).  If  we  did  not  know  the  factors  of  8a;2  +  22a:  +  15, 
we  could  find  them  as  follows  :  We  know  8  x^  +  22  a;+15 
=  (?a;+ ?)(?a:  +  ?).  Now  the  product  of  the  coefficients 
of  X  must  be  8,  and  the  product  of  the  last  two  terms 
must  be  15.  We  therefore  try  all  factors  that  meet  these 
conditions : 


2a:+5 

8  a: +  15 

4a;  +  3 

x-\-    1 

8a^+26a:+15 

8a^+23a;  +  15 

(wrong) 

(wrong) 

2a:-3 

2a;+3 

4a;-5 

4a;+5 

8a:2-22a^  +  15 

8  a:2  +  22  ar  +  15 

(wrong) 

(correct) 

This  may  take  a  long  time,  for  there  are  usually  many  possi- 
bilities that  occur  to  one  before  the  correct  combination  is  thought  of. 
If  we  notice  that  the  middle  term  is  the  one  that  conies  out  wrong  in 
case  of  a  wrong  guess,  we  need  try  only  it,  assuming  that  the  others 
will  be  right.  We  notice  that  the  middle  term  is  made  as  follows  : 
the  cross  products  are  those  marked  below  ;  their  sum  is  the  middle 
term : 

20  X  6  X        -  12  X  -  10  X  12  X  10  x 

We  therefore  try  just  these  cross  products,  and  decide  that  our 
choice  was  wrong  if  we  do  not  get  the  correct  middle  term. 

Ex.  1 .    Factor  4  ar^  -  13  x  + 10. 

Trying  various  pairs  of  numbers  whose  product  is  4  with  other 
pairs  whose  product  is  10,  we  finally  try  (4  x  —  5)(x  —  2)  ;  the  cross 
products  are  (-2)(4x)  =  — 8x  and  (— 5)(x)=  -  .'3x;  their  sum  is 
(— 8x)+ (— 5x)  =  —  13x,  which  is  the  correct  middle  term.  We 
therefore  multiply  (4  x  —  5)  by  (x  —  2)  to  test  our  result.  This 
gives  4x2  —  13  X  +  10;  hence,  this  pair  is  correct  and  the  required 
factors  are  (4  x  —  5)  and  (x  —  2). 


Hl_62]  FACTORS;   TYPE   FORMS  99 

Ex.2.   8  +  22x  +  15a:2  =  (2  +  3.r)(4-|-5x). 

Ex.  3.    8  m^  +  22  mn  -\- 15  n-  =  (2  m  +  3  n)  (4  m  -\-5n). 

The  letters  used  should  not  confuse  the  student. 

EXERCISES  XVI  :    CHAPTER   IV 

Perform  the  following  multiplications  : 

1.  (2a-3)(a-4).  4.    (5-x)(7  +  2x). 

2.  (3a  +  l)(5a-3).  5.    (x  +  5  y)(o x -  y). 

3.  (z-6k)(3z-\-k).  6.    (3x-2az){4tx  +  3az). 
Kesolve  into  factors : 

7.  8  2- +  8  2  —  6.  ^5    6/!L^Y_!^_12 

8.  3.^•2  +  a•2-52  2l  ^''^       ^' 

^     „  ^.    ,  16.    2a;*-x2-45. 

9_    vj fj vj^  fj-_ 

17.  12  .r^/ 4- 17  a;?/2  +  6  Z-. 

^  ^         ^  ^  18.  12  g'^  +  !7/i  —  6  /t^ 

11.  15-r-2r2.  ^g  2  aar' +  (4-3  a)a;- 6. 

12.  2  r^  +  r- 15.  20.  ahx"  +  {a +  h)tx -[-€-. 

13.  7  .r=  +  xy  -  6  ?/l  21.  12  ar^ -(4  6 -6)a;- 2  6. 

14.  6m2-13mH  +  6Ml  22.  4  e- +  9(30  e  +  479. 

62.  Other  Forms.  There  are  many  other  general  results 
which  might  be  given  here.  Among  them  we  mention 
the  following : 

(1)  (a  +  ft)(a2  -ah  +  h^)=a^  +  b^      See  Ex.  10,  p.  82. 

(2)  (a  -  5)(a2  +  ab+  V^)  =  a^  -  bK      See  Ex.  13,  p.  82. 

(3)  («2  _  b'^)(a^  +  b^)  =  a*  -  b\  or 

(3  a)   (a-?))(a  +  5)(a2  +  ^2)^a4_  J4. 

[This  is  really  an  application  of  §  59.] 

(4)  (a  +  by  =  rt3+  3  a%  +  3  ab"^  +  b\    See  Ex.  35,  p.  83. 

(5)  (a  +  ?>)4  =  a*  +  4  a^J  +  6  a262  +  4  ab^  +  b\ 

See  Ex.  36,  p.  83. 


100  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

(6)  (a  -  by  =  ffl3  _  3  a?b  +  Salfi-  h^     See  Ex.  38,  p.  83. 

(7)  (a  _  ^)4  =,a^_ia%  +  6  a%'^  -4:ab^-{-  b*. 

See  Ex.  39,  p.  83. 

These  and  many  others  may  be  found  by  the  student  Ijy 
multiplying  together  the  given  factors. 

As   before,  if   an   expression   can   be   put    into   a    form 

exactly  similar  to  one  of  these  answers,  its  factors  may 

be  found  by  comparison  with  the  factors  given  above  on 

the  left. 

EXERCISES   XVII:    CHAPTER   IV 

Factor : 

1.  p^-.S.        3.    x^-Sl.  5.    .16m*-n\        7.    ^)«-r/. 

2.  |r^  +  8.        4.    27-125a\        6.    r'^-f.  8.    a^'-b^ 
9.   ar'  +  3a;2  +  3.T  +  l.              14.   .r  -  30 ;)?  +  300  .v  -  1000. 

10.  f-6f  +  12t-  8.  15.  p'  -  I p'  +  h''-  /t P  +  8T- 

11.  l+4t  +  6f-\-4:fi+t\       16.  .r''  +  3  x-y  +  3  xf  +f  -  z\ 

12.  mV  +  216aj\  17.   d^ -Sa-h +oab- -  b'' -8c\ 

13.  {a-{-bf-\-(a-by.  18.   .r" -3 .r'- +  3.i- -  1  -  125 f. 
Perform,  by  factoring,  the  operations  indicated: 

a3  4.3a2^^3a^--.|,y  ^^    ^■4_  4  A"- +  (;  A:- -4  A-  +  1 

a'^2ab  +  b^        '  '        Jc" -3k' +  3k-l 

Perform  the  following  multiplications,  and  express  the  re- 
sult as  a  formula  in  each  case  : 

21.  {a'+ab-\-b-)(a'-ab  +  b-).       22.   (a' -b')(a'-{-b'). 

23.   (d,  +  />)•'.  24.    (a  -  bf.  25.    (rr  ±  by. 

63.  Factoring  by  Grouping.  Very  often  expressions 
may  be  factored  by  the  simple  process  of  grouping  the 
terms  together.  This  method,  which  is  ojie  of  the  most 
helpful  methods  known,  is  based  directly  on  the  laws  of 
p.  35.     llule  V,  p.  35 : 

a(b  +  c)=  ab  +  ac 


(iiMi.!]  FACTORS;    TYPE    FORMS  101 

is  ill  itself  an  exiimple  of  factoring,  and  is  the  chief  rule 
used  here;  this  has  already  been  used  extensively  in  mul- 
tiplication and  division.  The  same  method  has  been  used 
on  p.  79,  §  51,  in  finding  monomial  factors.  In  addition 
to  this,  expressions,  after  rearrangement,  may  be  compared 
w  ith  the  type  forms  given  above. 
Ex.  1.  ax  +  b>j  +  bx  -f  ay 

=  ax  +  bx  +  ay  +  by 
=  X  {a  +  b)  +  y{a  +  b)  =  {x  +  y)  (a  +  b) . 

Hence,  tlie  factors  of  ax-\-hy  +  bx  +  ay  are    (x  +  y)  and 

Check:  by  actually  multiplying  (a;  +  y)  and  (a +  5). 
Ex.2.  a2  +  4o6  +  462_9a,-2 

=  (rt-  4-  4  ah  +  4  6-)  —  9  3(? 

=  {a  +  2bJ-(?>xY 

=  [(a  +  2?>)+3.r][(a  +  26)-3a;]. 

Examples  similar  to  this  one  have  already  been  given 
in  the  preceding  lists. 

EXERCISES    XVIII:    CHAPTER   IV 

Factor  the  following  expressions  : 

1.  ab  —  a-  —  b-  +  (iJ).  9.   a^  —  ap  —  aq+pq. 

2.  xy  +  6  -\-2x  +  3y.  10.   .v^  —  ax'- -\- bcx  —  abc. 

3.  cr  —  1  —  a  +  a\  11.   rst  +  sH-  —  2  y^s  —  2  rsH. 

4.  ab  +  ac-bc-b-.  12.  Sz" -6zt +  lot^ -20zH. 

5.  ap  -h  3  qx  +  3  aq  +  px.  13.   t"  —  yf  —  art  +  a?y. 

6.  3  +  22;  — 6c  — 4  C2;.  14.   1^ +  at:- +  d- +  ht -^act  +  aJt. 

7.  ab^  —  abc  —  bx  -f-  ex.  15.   .t""*"  —  ax'"  —  6a;"  +  ab. 

8.  a^  — 4a^  +  3x--12.  16.   1—y  —  z+yz. 

17.  «-  —  o-  [  =  x^  +  a.j-  —  ax  —  o-]. 

18.  .r  +  2  o.i-  +  a-[  =  .jr'  +  a.r  +  o.r  +  «-]. 
19.   ic^  —  2  ox- -f  tr.  20.   x'^-a^ 


102  MULTlPLlCATIOiN    AND   DIVISION 

REVIEW  EXERCISES  XIX:   CHAPTER  IV 

Factor : 

1.  ar'+6a;-16.  10.  a;''-12a^+54a;--108a;+81. 

2.  x''-4:.  11.  27x^  +  Sy^ 

3.  p''  +  2pqi-q^-3(p  +  q)-10.  12.  a-  +  2ab  +  b--c^ 

4.  z^-64:.  13.  3^;*- 8:2- -35. 

5.  4:  a' -  20  ax  +  25  X-.  14.  9  +  18a-7al    V 

6.  aV-3a26V+3a6^j;-6^  15.  l-'y-156y-. 

7.  a;* -16.  16.  81  i^- 256. 

8.  jj^  -  10  jyq  +  25.  ^  17.  10-a;-3a;2. 

§;.  10  -  C2  -  60  c2;2^  /  18.  JJ*  -  81  ^^ 

19.  63  =  64-1;  1001  =  1000  +  1;  65  =  64  +  1;  77  =  81-4. 

20.  m*  —  4:  m^n  +  6  m-u^  —  4  ?«,?i^  +  n'^—  16 p*. 

21.  fe^  -  12  A:^  +  54  fc2  -  108  A;  -  175  =  (k  -  Sy  -  256.     [Factor.] 

22.  a^  —  Jf  =  (o^  -  W)  (a'  +  V").      [Factor  these  factors.] 

23.  a'  +  cv'h'  +  h'={a'  +  2a'l)-  +  W)~crb-.      [Factor.] 

24.  «•=  -h^=  (a^  -  h-) {a*  +  a-^^  +  b^).      [Factor  these  factors.] 

25.  a»  -  6*  =  (a*  -  b*)(a*  +  6*).      [Factor  these  factors.] 

26.  a^"  -  &'"  =  («'  -  &')(«'  +  l>').      [Factor  tliese  factors.] 

27.  a^o  _  ^10  ^  c„:i  _  f/.^ (^jH  _^  ^j6^2  _,_  ^,4i,i  ^  ,x-U-'  +  b^) .      [  Factor.] 

28.  a»  +  «"?>-  +  <<'^/  +  crU'  +  ?>^      [Make  use  of  Exs.  20,  27.] 

29.  a-  —2ab-\-  b-  —  :r  +  2 xy  —  y: 

30.  a"  -  3  tv'b  +  3  al/  -W-  x"  +  3  ar^*/  -  3  a-/  +  f. 

31.  ^.^-'  -  3  ^.T  +  2  .4  +  Bx^  -.3Bx-\-2B. 

32.  y /•  +  2  y^  +  7  77>  +  14  ^>/  +  24  /  + 12  r. 

33.  l  —  r  —  s—t  +  rt-\-  rs  +  «^  —  rst. 


PART  IV.    APPLICATIONS.    ENGLISH  TRANSLATED  INTO 

ALGEBRA 

64.  Expression  of  English  in  Formulas.  The  preceding 
chapters  have  given  practice  in  writing  formulas  in  the 
phice  of  English. 

Thus,  instead  of  the  rule  :  "  The  square  of  the  sum  of  two  terms  is 
equal  to  the  sijuareof  the  first  term,  plus  twice  the  product  of  the  two 
terms,  plus  the  scjuare  of  the  second  term,"  we  merely  write 

(a;  +  ?/  )2  =  x'^  -\-  '2  xy  +  y'^, 

which  really  says  the  same  thing  in  much  shorter  form. 

Again,  instead  of  saying,  "  The  volume  of  a  box  in  cubic  inches 
is  the  product  of  its  width  times  its  length  times  its  height,  each 
measured  in  inches,"  we  may  simply  say 

V  =  11}  •  I  •  h, 

where  v  means  volume  in  cubic  inches,  and  lo,  I,  h,  stand  for  the  width, 
length,  and  height,  respectively,  measured  in  inches. 

In  nian}^  cases  the  student  will  find  it  easier  to  write 
down  the  formulas  if  he  first  tries  the  problem  with  known 
numerical  values.  In  doing  so  it  is  advisable  not  to  per- 
form the  additions,  multiplications,  etc.,  but  merely  to 
indicate  them  as  a  guide  in  the  case  when  no  numbers 
are  actually  given.  In  this  way  the  structure  of  the  ex- 
ample is  seen  with  simple  numbers,  and  can  be  followed 
afterwards  in  the  use  of  letters  for  unknown  numbers. 

Consider  the  problem: 

Ex.  1.  A  till  box  is  to  be  made  from  a  square  piece  of  tin 
by  cutting  square  pieces  out  of  the  corners  and  then  folding 
up  the  flaps.  Find  tlie  size  of  the  piece  of  tin  which  must  be 
used  to  make  a  box  4  in.  high  that  shall  contain  100  cu.  in. 
volume. 

Let  us  first  become  familiar  with  the  problem  by  trying  several 
numbers.     Suppose    the    original   plate  were  18    in.    square   (in  fig- 

103 


104  MULTIPLlCATIOiN    AND   DIVISION  [Cii.  IX 

lire,  AB  =  EC  =  18  in.),  and  we  cut  out  the  shaded  corners,  each 
4  in.  square.  Then  HE  =  18  in.  —  2  x  4  in.  =  10  in.  Likewise 
FE  =  10  ill.     If  we  now  fold  up  the  flaps,  we 


yyx/Mi  Y///////\^    gg^  ^  -^^^  whose  bottom  is  10  in.  square  and 


G  F 


H  E, 


whose  height  is  4  in.     The  volume  of  this  box 
is  10  in.  X  10  in.  x  4  in.  =  400  cu.  in. 

It  is  now  easier  to  tiy  the  given  problem. 
To  start  with,  we  do  not  know  how  long  AB 
must  be.     Let  us  call  its  length  /  in.  and  then 
-°    try  to  find  Z.     If  we   cut  the  corners  out   as 
*'^"-  ^^-  before,  we  shall  have  HE  =  /  in.  -  2  x  4  in. 

=  (/ —  8)  in.  just,  as  before.  Likewise  FE  =  {I  —  8)  in.  Hence, 
our  box  will  have  its  bottom  (I  —  8)  in.  square.  Its  height  will 
be  4  in.  Its  volume  is  then  (I  —  8)  in.  x  (Z  —  8)  in.  x  4  in.  = 
4(/  —  8)''^  cu.  in.     If  this  volume  is  to  be  100  cu.  in.,  we  shall  have 

4(Z-8)2=  100. 
Divide  both  sides  by  4  :  (Z  -  8)2  =  25, 

hence,  (Z  —  8)  =  5  or  else  —  5, 

since  (5)2  =  25,  and  also  (- 5)2  =  25.* 

Adding  8  to  each  side,  we  get  Z  =  8  +  5  or  else  8—5. 

It  follows  that  Z  =  13  or  else  I  =  3. 

The  correct  result  is  Z  =  13,  for  Z  =  3  would  not  really  do  at  all ;  we  could 
not  cut  corners  4  in.  square  out  of  a  piece  of  tin  3  in.  square. 

As  in  this  problem  the  student  must  always  be  careful 
to  see  which,  if  any^  of  several  possible  answers  are  the 
correct  ones,  for  there  may  be  answers  that  cannot  pos- 
sibly mean  anything,  as  is  seen  above. 

Ex.  2.  If  a  printed  book  is  to  have  a  margin  2  in.  wide, 
how  large  must  the  pages  be  cut  if  there  is  to  be  70  sq.  in.  of 
print  on  them  and  the  pages  are  to  be  twice  as  long  as  they 
are  wide  ? 

*  There  are  no  other  numbers  whose  square  is  25,  for  a  positive  num- 
ber less  than  5  would  give  a  square  that  is  too  small ;  a  positive  number 
greater  than  5  would  give  a  square  that  is  too  large  ;  and  similarly  no 
negative  number  except  —  5  would  produce  precisely  25.  It  is  very  im- 
portant, in  all  problems,  to  make  sure  that  the  answers  found  are  all  that 
exist.  If  this  is  not  done,  some  answer  —  perhaps  the  most  important  — 
may  be  overlooked. 


6i-r)5j 


KXPIIESSJOX    IN    FORMULAS 


105 


Fig.  19. 


Let  us  first  become  familiar  witli  the  problem.  Suppose  the  page 
is  10  in.  X  "20  in.  (twice  as  long  as  wide). 

Then  AB  =  10  in.,  BC  =  2  x  10  in.  =  20  in. 
If  the  margin  (on  each  edge)  is  2  in.  wide,  then 
EF  =  10  in.  -  2  X  2  in.  =  6  in., 
and  FG  =  20  in.  -2x2  in.  =  16  in. 

The  printed  portion  is  then  (6  x  16)  sq.  in.  =  96  sq. 
in.  Suppose  now  that  the  width  AB  is  not  known 
in  advance;  call  it  win.     Then 

AB  =  w  in.     BC  =  2  x  w  in.  =  2  w  in. 
If  the  margin  (on  each  edge)  is  2  in.  wide,  then 

EF={w  -  2)x  2  in.  =  {w  -  4)  in., 
and  FG  =  [2  ic  -2x2]  in.  =  (2  w  -  4)  in. 

Hence,  the  printed  portion  is 

(w  —  4)  (2  w  —  4)  sq.  in.,  or  (2n'^  ~  12  w  +  16)  sq.  in. 
But  the  printed  portion  is  to  be  70  sq.  in.,  hence, 
2  w2  -  12  w  +  16  =  70. 

Divide  each  side  bv  2  : 

w-  -6io  +  S  =  35. 

Subtract  35  from  each  side : 

iv'^  —  6  w  —  27  =  0. 

But  this  is  (lo  +  3)(?o  -  9)  =  0   (§60). 

Hence,  ?<j  +  3  =  0,  or  else  iv  —  9  —  0, 

for  the  product  (w  +  3)(»'  —  9)  would  not  be  zero  unless  one  of  the 
factors  were  zero. 

If  w  +  3  —  0,  then  w  =  —  3 ;  this  is  meaningless,  for  a  piece  of 
paper  cannot  be  —  3  in.  wide. 

If  w  —  9  =  0,  then  rv  =  9  ;  and  this  is  soon  seen  to  be  correct,  for  if 
w  =  9,  the  actual  printed  area  is  5  in.  wide  and  14  in.  long;  its  area  is 
therefore  (5  x  14)  sq.  in.  =  70  sq.  in.,  as  was  required. 

65.  Simple  Changes  in  Equations.  We  have  made  several 
changes  in  the  equations  above  which  we  shall  now  review. 

Thus,  we  had  in  example  1,  4 (Z— 8)2=100  and  we 
divided  each  side  hy  4.  The  justification  for  this  is  that 
4(Z—  8)2  is  the  same  number  as  100;  hence,^  of  either  of 
them  is  equal  to  \  of  the  other.     (Compare  §  35,  p.  55.) 


106  MULTIPLICATION    AND   DIVISION  [Cii.  IV 

Likewise,  we  multiplied  both  sides  by  the  same  number, 
added  the  same  number  to  both  sides,  subtracted  the  same 
number  from  both  sides.     The  argument  is  the  same. 

We  see  that  in  any  equation,  if  it  is  a  true  equation, 
the  two  sides  stand  for  the  same  number;  hence: 

Given  any  equation,  ice  may 
I.    Add  the  same  nu^nher  to  each  side. 
II.    Subtract  the  same  number  from  each  side. 

III.  Multiply  each  side  by  the  same  number.* 

IV.  Divide  each  side  by  the  same  7iumber  except  zero.f 

V.  Perform  the  same  ojyeration  (of  any  kind)  on  the 
whole  of  each  side,  if  the  residt  is  known  to  be  a  single  num- 
ber in  each  case.^  (Take  care  to  perform  the  operation  on 
the  whole  of  each  side,  not  on  a  part  of  it.) 

These  principles  have  been  used  frequently;  the  first 
four  were  stated  in  another  form  in  §  35,  p.  55. 

66.  Product  Equal  to  Zero.  Another  principle  is  the 
one  used  in  example  2,  p.  104.  Since  (iv  +  3)(w  —  9)  =  0, 
we  knew  that  either  (w+o)=U,  or  tluit  (w  — i))  =  0, 
for  otherwise  the  product  of  the  two  could  not  be  zero. 

*  If  both  sides  are  multiplied  by  zero,  the  new  equation  is  0  =  0,  which 
is  correct,  but  not  useful.  Avoid  multiplying  both  sides  by  zero,  and  test 
any  expression  used  as  a  multiplier  to  see  if  it  is  zero  for  the  values  finally 
found. 

t  In  dividing,  great  care  is  sometimes  necessary  to  avoid  dividing  by 
zero.     Notice  that  no  number  can  be  divided  by  zero.     (See  p.  75.) 

t  If  the  result  of  the  operation  performed  is  not  a  single  number, 
great  care  is  necessary.  Thus,  we  said,  in  example  1,  if  (Z  —  8)^  =  25,  we 
know  that  Z  —  8  is  either  +  5  or  —  5.  To  get  this  we  take  the  square 
root  of  each  side.  But  25  has  tico  square  roots,  since  (+5)-=(  +  5)x 
(+  5)  =  25  and  (—  6)-  =  (—  5)  x  (—  5)  =  25.  Hence,  we  cannot  be 
sure  which  one  of  these  is  equal  to  (1  —  8).  Nevertheless,  we  may  make 
a  perfectly  definite  statement  even  in  this  case  ;  namely,  "  (I  —  8)  is 
equal  ciUu-r  to  5  or  else  to  —  5." 


65-00]  EXPRESSION    I\    FORMULAS  107 

This  principle  is  simply  that  two  numbers,  neither  of 
which  is  zero,  have  a  product  which  is  not  zero.  This 
truth  is  quite  evident  after  a  little  thought,  and  we  shall 
state  it  as  follows : 

VI.  //"  the  j^roduct  of  two  factors  is  zero,  then  at  Ie((,st 
one  of  the  factors  is  zero. 

Let  us  solve  additional  problems  to  illustrate  these  prin- 
ciples : 

Ex.  1.  Each  member  of  a  certain  family  gave  each  of  the 
others  a  present,  at  Christnms,  wliicli  cost  50  cents.  If  the 
total  sjient  by  the  family  was  $10,  how  many  persons  are 
there  in  the  family  ? 

To  become  familiar  with  the  problem,  suppose  there  had  been  6 
members.  Then  each  one  would  have  given  5  presents,  so  that  all 
together  6x5,  or  30,  presents  would  have  been  given. 

Since  the  number  in  the  family,  is  not  really  known,  let  us  call 
it  n.     Then  each  member  gave  n  —  1  presents,  so  that  all  together 
there  were  n(n  —  1)  presents  given.     If  each  cost  50  cents,  the  cost 
in  cents  was  50/i(;j  —  1).     But  this  is  known  to  be  1000  cents: 
50n(n  -  1)=  1000. 

Divide  both  sides  by  50  : 

n(7i  -  1)=  :20. 

Multiply  n  —  1  by  ?!,  and  subtract  20  from  each  side, 
„2  _  „  _  20  =  0, 
(n  +  4)  (n  -  5)  =  0. 
Hence,  one  or  the  other  of  these  factors  is  zero : 

71  +  4  =  0,  or  »i  —  5  =  0 ; 
that  is,  subtracting  4  from  each  side  or  adding  5  to  each  side  gives, 

n  =  —  4,  or  n  =  5. 
Now  n  =  —  4  is  meaningless  in  this  problem  ;    therefore  n  =  5 :   there 
are  5  persons  in  the  family.     To  check  this  result,  we  notice  that  there 
would  then  be  5  x  4  =  20  presents,  which,  at  50  cents  each,  would 
cost  1 10.     This  check  is  complete. 

Many  practical  problems  arise  in  computing  the  effect 
of  errors  in  nieasnrements.      In  sucli  problems  the  error 


108  MULTIPLICATION   AND   DIVISION  [C'n.  IV 

in  measurement  is  usually  counted  positive  when  the  meas- 
urement is  too  large  ;  if  the  measurement  is  too  small,  the 
error  is  said  to  be  negative.  The  following  examples  will 
illustrate  the  calculations  of  such  error  effects. 

Ex.  2.  Let  e  denote  the  error  (in  inches)  made  in  the  meas- 
urement of  the  side  of  a  square  whose  side  is  really  10  ft.  long; 
and  let  E  denote  the  error  (in  square  inches)  in  the  computed 
value  of  the  area  of  the  square.  Express  E  in  terms  of  e.  If 
e  =  3  (in.),  find  E.  U  E  =  484  (sq.  in.),  find  e.  U  E  =  -  119| 
(sq.  in.)  {i.e.  the  computed  area  is  119f  sq.  in.  too  sro.all),  find  e. 

The  side  is  really  10  ft.,  or  120  in.  long.  If  the  error  is  e,  the  nieas- 
m-ement  is  120  +  e  (in  inches).  Hence,  the  computed  area  is  (120  +e)'' 
(in  square  inches).  Since  the  real  area  is  (120)'^  (in  square  inches), 
the  error  E  in  the  computed  area  is 

E  =  (120  +  ey  -  (120)2  ^  g-2  ^  240  e. 

If  e  =  3  (in  inches),  the  value  of  E  is  found  by  putting  3  in  place 
of  e  in  the  preceding  equation ;  and  we  find 

£  =  32  +  (240)(3)  =  9  +  720  =  729; 
hence,  the  effect  of  an  error  of  3  in.  in  measuring  the  length  of  the 
side  of  the  square  causes  an  error  of  729  sq.  in.  in  the  computed  area, 
or  about  5  sq.  ft. 

If  E  =  484  (in  square  inches),  we  have 

e2  +  240e  =  484,    or   e^  +  240e  -  484  =  0; 
whence,  (e  -  2)  (e  +  242  )  =  0 ; 

and  either  e  -  2  =  0,    or   e  +  242  =  0 ; 

that  is,  either  e  =  2,   or   e  =  —  242. 

The  answer  e  =  —  242  is  unreasonably  large,  since  no  one  would  con- 
ceivably make  an  error  of  the  amount  of  242  in.  in  measuring  a  length 
of  10  ft.  The  answer  e  =  2  is  evidently  the  only  one  to  which  we 
need  pay  attention  ;  it  means  that  the  error  of  484  sq.  in.  in  the  com- 
puted area  would  be  caused  by  an  error  of  2  in.  in  the  measurement 
of  the  side,  the  measurement  being  2  in.  too  long,  since  e  is  positive. 

It  E  =  —  119|  (in  square  inches),  we  have 

e2  +  240  e  =  -  119|,    or   e^  +  240  e  +  119^  =  0, 
or,  multiplying  by  4,  we  get 

4  ^2  +  960  e  +  479  =  0. 


.i(i-(i7J  EXFKKbbiUN    IN    FORMULAS  101) 

Since  479  has  no  factors  except  itself  and  1,  we  would  soon  try  the 
factors  (2e  +  1)  and  (2  e  +  47!») ;  these  are  correct,  as  will  be  seen  by 
nui'tiplying  them  together.     Since  their  product  is  zero,  we  have 

either  2  6+1=0,   or   2  e  +  479  =  0, 
that  is,  eitlier  e  =  -  l,   or   e  =  -  239^. 

The  answer  e  =  —  \  means  that  the  error  made  in  measuring  the 
side  was  negatice,  i.e.  that  the  measurement  was  too  short,  and  the 
amount  of  the  error  was  |  in.  The  other  answer  e  -  —  239^  is  un- 
reasonably large,  and  we  need  not  consider  it. 

Check:  If  e  =  -  ^,  the  measured  length  was  119^  in.;  hence  the 
computed  area  w^as  (119^)2  (sq.  in.)  or  14,280^  sq.  in.  The  real  area 
being  14,400  sq.  in.,  the  error  E  was  -  119f  sq.  in.  (correct). 

In  the  i^receding  problem,  the  answer  e  =  —  J  has  a  real  meaning, 
though  it  is  negative.  In  problems  about  actual  things,  always  notice 
carefully  whether  or  not  a  negative  answer  can  be  interpreted.  Notice 
also  whenever  an  answer  is  unreasonably  large,  or  when  it  is  unreason- 
able for  any  other  cause,  and  be  careful  to  say  whij  any  of  several  pos- 
sible answers  is  discarded.     See  also  the  footnote  on  p.  104. 

67.  Solution  of  Problems.  The  following  exercises  will 
illustrate  the  uses  of  factoring  and  of  multiplication  and 
division  in  simple  cases.  They  may  all  be  solved  by  the 
principles  of  §§  G5-66  ;  and  the  answers  found  will  be 
correct  if  these  principles  are  carefully  followed. 

The  equations  solved  above  are  quadratic  equations; 
we  shall  study  such  equations  in  more  detail  in  Chapter 
VIII,  p.  203. 

Apparent  answers  maybe  found  which  do  not  fulfill  the  conditions 
of  the  given  problem.  Thus,  the  equation  (1)  x  —  3  =  0  has  only 
one  answer,  x  =  3.  But  if  both  sides  of  equation  (1)  are  multiplied 
by  a:  -  2,  the  resulting  equation  {x  -  2)(x  -  3)  =  0  has  two  possible 
answers,  x  =  2  and  x  =  3,  the  first  of  which,  x  =  2,  is  not  a  possible 
answer  for  equation  (1).  Xo  such  false  answer  can  arise  if  both  sides 
are  multiplied  by  a  number,  as  required  in  III,  §  6.5,  if  that  number  is 
not  zero. 

In  order  to  be  sure  that  no  mistakes  have  been  made  in  the  work, 
as  well  as  to  avoid  the  possible  introduction  of  false  answers,  each 
answer  should  be  checked,  as  above,  by  actually  trying  it  in  the  given 
problem.     Such  a  clieck  is  complete. 


110  MULTIPLICATION   AND   DIVISION  [Cii.  IV 

EXERCISES   XX:    CHAPTER   IV 

1.    In  example  1,  p.  103,  we  found  that  the  given  problem 

led  to  the  formula  v  =  4 (?  —  8)1  Find  v  when  1  =  8;  when 
1  =  9;  10  ;  11 ;  13.  Find  v  when  Z  =  0;l;3;5;7;-l;-5. 
]\Iake  a  table  of  these  values  of  v  and  I,  and  plot  the  graph  for 
this  equation.  Does  the  shape  of  the  graph  resemble  any  you 
have  drawn  before? 

When  V  =  100,  determine  I  from  the  graph.  This  procedure 
is  called  the  graphical  sohition  of  the  problem.  Compare  your 
results  with  the  results  on  p.  104. 

2..  When  v  =  12\,  l  =  ?  What  is  v  when  1  =  2?  when 
Z  =  3i?     Solve  both  graphically  and  algebraically. 

For  any  value  of  I,  how  many  values  of  v  are  there?  For 
any  value  of  v,  how  many  values  of  I  are  there?  In  the  case 
of  two  results  such  as  v  =  100,  I  =  13  or  3,  how  is  the  correct 
answer  to  be  distinguished  from  the  incorrect  one  in  the 
figure  ? 

3.  In  example  2,  p.  104,  the  printed  portion  p  =  2  ro^  — 12  w 
+  16,  if  w  =  width.  Find  piiw  =  0;  1 ;  2  ;  5  ;  10  ;  -  1 ;  -  5  ; 
—  10.  IMake  a  table  and  draw  the  graph.  Solve  graphically 
for  w  if  p  =  70,  and  compare  your  results  with  those  on  p.  105. 
How  many  values  of  to  are  there?  Point  out  why  one  is 
impossilde  as  a  true  solution. 

4.  A  certain  rectangular  yard  is  known  to  contain  G50 
square  feet;  it  is  found  that  a  ir)0-foot  rope  exactly  surrounds 
the  fence.  What  are  the  length  and  breadth  of  the  yard? 
Solve  by  letting  I  =  length. 

5.  If  A  denotes  the  area  of  a  field,  as  in  Ex.  4,  what  rela- 
tion connects  A  and  I?  Draw  the  graph.  Solve  graphically 
for  I  if  A  =650.  What  can  you  say  of  the  two  values  obtained 
for  I  in  this  case? 

6.  For  what  values  of  /  in  Ex.  5  is  ^1  =  0?  What  is  the 
space  between  these  values?  If  the  number  I  represents 
the  length  of  the  field,  what  distance  in  the  figure  represents 


(17]  EXPRESSION    IN    FORMULAS  111 

the  breadth  of  the  field?  Point  out  on  the  figure  the  greatest 
possible  value  of  the  area  A.  For  this  value  of  A,  what  are 
the  length  and  breadth  of  the  field  ? 

7.  By  what  amount  must  the  length  and  breadth  of  a  rec- 
tangular plot  of  ground  300  ft.  long  by  50  ft.  wide  be  equally 
increased,  in  order  that  the  area  may  be  increased  by  3G00 
sq.  ft.  ? 

8.  Represent  by  a  graph  the  increase  in  area  of  the  plot 
of  ground  corresponding  to  equal  increase  of  length  and  breadth. 
Solve  the  problem  graphically.  Interpret  negative  values. 
What  equal  decrease  of  length  and  breadth  will  reduce  the 
area  5025  sq.  ft.  ? 

9.  What  error  (e)  in  measuring  the  side  would  cause  an 
error  (E)  of  241  sq.  in.,  the  computed  area  of  the  square  men- 
tioned in  Example  2,  §  66  ?  If  e  =  5,  find  E ;  if  e  =  -  3,  find 
E;   HE  =  -239,  find  e. 

10.  The  sum  of  the  squares  of  two  consecutive  integers  is 
113.     What  are  the  integers  ? 

Have  both  positive  and  negative  results  a  meaning  in  this 
problem  ? 

11.  "Think  of  a  number;  add  3  to  it;  nudtiply  the  result 
by  2 ;  from  this  result  subtract  5,  then  multiply  by  the  num- 
ber you  first  thought  of.  WHiat  is  your  result?"  If  the 
result  given  is  36,  what  is  the  number  thought  of  ?  If  the  re- 
sult given  is  21,  what  is  the  number  thought  of? 

12.  Show  that  the  difference  of  the  squares  of  two  consecu- 
tive integers  n  and  n  -f  1  must  be  an  odd  number.  If  the  dif- 
ference of  the  squares  is  given,  how  may  the  numbers  be  found  ? 

Determine  two  consecutive  integers  such  that  the  difference 
of  their  squares  is  9,  15,  33. 

13.  For  the  frame  of  a  picture  10  by  15  in.,  116  sq.  in.  of 
material  are  available.  How  wide  may  the  frame  be  made  if 
the  frame  is  the  same  width  on  all  sides,  and  if  it  exactly 
meets  the  picture  ? 

Solve  graphically,  also. 


112  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

14.  The  interest  on  a  sum,  p,  of  money  at  a  given  rate,  r,  for 
a  given  time,  t,  is  i=prt.     What  is  the  amount,  A  ? 

What  is  the  amount,  A,  for  one  year  ?  What  is  the  amount, 
A',  at  annually  compounded  interest,  for  two  years? 

Plot  the  annually  compounded  interest  for  two  years  on 
$200  at  various  rates.  (Do  not  start  the  plotting  of  A'  from 
0,  but  (say)  from  $200.  Choose  rates  from  0%  to  12%  at  in- 
tervals of  about  i%.)  For  what  rate  will  A'  be  $220.50? 
Solve  first  graphically,  then  by  the  equation. 

15.  A  rectangular  frame  is  so  constructed  that  by  means  of 
a  slide  its  width  may  be  altered.  When  it  is  set  at  the  width 
5  centimeters,  a  rod  13  centimeters  long  just  fits  as  diagonal. 
How  much  must  the  width  be  increased  so  that  a  rod  15  centi- 
meters long  may  fit  as  diagonal  ? 

16.  A  rectangular  plot  of  ground  60  feet  by  14  feet  is  to  be 
doubled  in  area  by  equal  increase  of  length  and  breadth. 
Draw  a  picture  illustrating  in  general  the  alteration  in  area 
corresponding  to  equal  alteration  of  length  and  breadth.  Find 
the  alteration  necessary,  both  by  use  of  the  graph  and  by 
solution  of  the  equation. 

17.  Let  the  error  in  the  measurement  of  the  side  of  a 
square  whose  side  is  really  15  ft.  long  be  denoted  by  e  (in 
inches),  and  the  error  in  the  computed  area  by  E  (in  square 
inches).  Express  E  in  terms  of  e.  If  e  =  1,  find  E;  if  e  =  2, 
find  E;  if  e  =  3,  find  E ;  etc. ;  if  e  =  —  1,  find  E ;  etc.  Draw 
the  graph,  taking  one  small  space  on  the  horizontal  line  to- 
mean  1  in.  in  values  of  e,  and  one  small  space  on  the  vertical 
axis  to  mean  1  square  foot  in  values  of  E,  i.e.  144  sq.  in. 
Find  e  if  ^  =  361 ;  if  ^  =  724 ;  if  ^  =  -  359. 

18.  Let  the  error  in  measuring  the  radius  of  a  circle  whose 
radius  is  really  7  ft.  long  be  denoted  by  e  (in  inches),  and  the 
error  in  the  computed  area  by  E  (in  square  feet).  Taking 
TT  =  3f  (see  table  at  back  of  book),  show  that  E  =  528  e  +  3|  e'^. 
Find  E  for  various  values  of  e;  plot  the  figure;  find  e  if 
E=-  524f . 


67] 


REVIEW 


113 


REVIEW   EXERCISES   XXI:   CHAPTER   IV 

Perform  the  indicated  multiplications  ;  check  : 

1.  (a^  -  3  d'h  +  5ab''-  b^)  (a-  -  ah  +  6"). 

2.  (16^'2-4A;  +  l)(16A;-  +  4fc  +  l). 

3.  {l-a-b)(l  +  a-^-h  +  d'  +  2ab  +  b^. 

4.  {a  +  2b  +  3c){b  +  2c  +  3a)(c  +  2a  +  3b). 

5.  {y-z)(z-x)(x-y). 

6.  (a*  +  a^  +  «^+a  +  l)(a-l). 

7.  (m3"  +  m-''  +  m"  +  l)(m"  — 1). 

8.  (ax- +  bx  +  c)(lx- +  mx  +  n). 

9.  —  a''b''c''  X  a^^c"  x  —  a''&''c\ 

10.    (10-3)(5  +  7);  (6-9)(8-10);  (7-ll)(3  +  6). 

Perform  the  indicated  divisions  (inexact  divisions  indicated 
by  *,  as  before) ;  check  each  exercise  : 


11. 


13. 


^5  _  2"  4-  8  2=^  - 11  z-  +  21z 


12. 


a^  4-  (g  +  6)  a;  +  «& 


z^—2z^  +  3z  x  +  a 

—  a;«  +  3  a^  +  2 a;^  -  a;=^-  14  ar^ -  9  g;  + 18 
a;^  — 5a;  +  6 


a^  —  (g  +  6  +  c)  a.-^  +  (be  +  ca  +  ab)x  —  abc 
x^  —  {a  +  b)x  +  ab 


*15. 


18. 


19, 


*20. 


x^+px  +  q 


*16. 


a^  +  j)x-  +  qx  +  r  _ 


17. 


■b" 


X —a  x—a  a+b 

y6  +  Q  fz  + 15  v'z'  +  20  y V  + 15  y"^*  4-  6  yz^  +  z^  _ 

y-  +  2yz  +  z- 

2  m^  +  3  rJ  +  6  m^n  -  3  mn""  +  m'><'  +  3  ^^'n'  +  3  mh}\ 
~  2  wi^  +  4  m^n  —  6  mn^  +  3  n^ 

a;^  +  3ar5  +  10a-2+14a;  +  10  *  gi.    P' +  ^ . 


af+2x 


p-q 


114  MULTIPLICATION   AND   DIVISION  [Ch.  IV 

Factor  tlie  following  expressions  : 

22.  rc-  +  10a;  +  24.  29.  Gx'  +  x-SB. 

23.  x^- 10  a- +  24.  30.  63  A' +  AB- 12  B\ 

24.  a;2  +  10.^■-24.  31.  z*  -  625. 

25.  a.-2  -  10  a;  -  24.  32.  a^  +  5d'  +  9. 

26.  T'  +  2ax  +  {a'-b'^.  33.  r* +  10,^  +  9.     • 

27.  a" -61  34.  r^-z^ 

28.  2  a.-^  +  2  ^- -  40.  35.  x^-if". 

36.  ^/  —  5  ^/g  +  10  j/q-  -  10  p'Y  +  5  /Ky"  —  (/'. 

37.  r^  -  12  7'"^  +  54  v-^  -  108  y  +  81 . 

38.  (>*"  +  3  71-  +  4)  =  (//"  +  4  7*2  +  4)  —  ?tl 

39.  a^.r'"  —  (a-  +  l'>-).^•"'2/  +  f'6^-. 

40.  x'^—lAx^ +  49 —  y-+2  yz  —  z^. 

In  the  following  exercises  solve  both  from  the  graph  an<] 
from  the  eqnation,  unless  otherwise  directed : 

41.  A  page  is  to  have  a  margin  of  1  inch,  and  to  contain 
35  square  inches  of  printing.  How  large  must  the  page  be  if 
the  length  is  to  exceed  the  width  by  2  inches  ? 

42.  How  wide  a  path  may  be  laid  out  inside  the  margin  of 
a  park  area  100  x  250  feet,  in  order  that  the  space  taken  up  by 
the  path  may  be  3400  square  feet  ? 

43.  Such  a  problem  as  Ex.  42  would  arise  as  follows :  how 
wide  a  path  will  take  up  an  area  not  greater  than  4000  square 
feet,  approximately.     Solve  this  problem  graphically. 

Note  that  we  are  not  yet  in  a  position  to  solve  the  problem 
even  approximately  by  means  of  the  equation. 

44.  For  a  kindergarten  gift,  a  rectangular  box  with  a  square 
top  is  to  be  constructed  5  inches  high.  If  exactly  2  square 
feet  of  colored  paper  is  used  to  cover  the  box,  including  the 
top  and  the  bottom,  what  must  be  the  length  of  one  side  ? 


fiT]  REVIEW  11  o 

45.  In  the  preceding  exercise,  it  is  desired  that  not  more 
than  5  square  feet  of  paper  be  used  to  cover  the  box.  How 
large  may  the  side  be  chosen  ?     Solve  only  graphically. 

46.  "  Think  of  a  number,  double  it,  add  three,  and  square 
the  result ;  from  this  result  subtract  seven,  and  halve  the  re- 
mainder."    Tlie  result  is  21  ;  what  was  the  number  chosen  ? 

47.  Construct  a  puzzle  similar  to  the  above,  leading  to  an 
easy  equation,  involving  the  square  of  the  unknown  number; 
propound  it  to  a  friend  and  find  the  number  chosen  by  him. 

48.  Find  two  numbers  whose  sum  is  14  and  whose  product 
is  48. 

49.  The  hypotenuse  of  a  certain  right-angled  triangle  is 
ten  feet  long;  a  piece  of  string  which  is  exactly  one  fourth 
as  long  as  one  of  the  perpendicular  sides  is  found  to  be  exactly 
one  third  as  long  as  the  other.  How  long  is  the  string,  and 
how  long  is  each  of  the  perpendicular  sides  ?    (See  Tables.) 

50.  What  radius  must  a  cone  whose  slant  height  is  5  inches 
have  in  order  to  be  covered  by  14  tt  square  inches  of  tinfoil  ? 

51.  What  is  the  error  E  (in  square  inches)  in  the  computed 
value  of  the  area  of  a  square  50  ft.  long,  caused  by  an  error  e  (in 
inches),  in  measuring  the  side  ?  Plot  the  graph  as  in  Ex.  17, 
p.  112.     Find  e\iE  =  2404  ;  \i  E  =  -3591. 

52.  How  carefully  must  the  side  of  the  square  in  Ex.  51  be 
measured  in  order  to  be  sure  that  the  error  in  the  computed 
area  is  not  more  than  2404  sq.  in.  ?  How  carefully  must  each 
mark  be  made  if  the  side  is  measured  with  a  foot  rule  in  the 
ordinary  way  ? 

53.  Answer  similar  questions  for  E  =  6025  in  Exs.  51  and 
52.  Is  it  reasonable  to  suppose  that  the  area  can  be  computed 
(from  measurement  with  a  foot  rule)  to  within  10  sq.  ft.  ? 

54.  How  carefully  must  one  measure  the  radius  of  a  circle 
whose  radius  is  21  ft.  in  order  that  the  computed  area  may  be 
correct  to  within  7951  sq.  in.  ?  Is  it  reasonable  to  suppose 
that  the  area  can  be  computed  (from  measurement  of  the 
radius  with  a  foot  rule)  to  within  5  sq.  ft.?  (See  Ex.  18,  p.  112.) 


116  MULTIPLICATION    AND   DIVISION 


SUMMARY    OF    CHAPTER   IV:    MULTIPLICATION   AND   DIVI- 
SION;  FACTORING;    APPLICATIONS;   pp.  67-115 

Pakt  I.     Multiplication  and  Division  of  Numhkrs  and  Mono- 
mials, pp.  67-77. 

Dejinition  of  Multiplication  in  General:  fundamental  properties  to  he 
preserved.  §  40,  pp.  67-68. 

Product  of  Negative  and  Positive  :  product  of  amounts  preceded  by 
-  sign.     Exercises  T.  §  II,  pp.  68-70. 

Product  of  Two  Negatives :  product  of  amounts  preceded  by  +  sign. 
Rule  of  Signs:  like  signs  give  +,  uidike  give  —  .     Exercises  II. 

§  42,  pp.  70-71. 
Multiplication  of  Monomials  :  rearrangement  of  factors.     Exercises 
HI.  §  43,  pp.  71-72. 

Multiplication  of  Powers  of  same  letter;  add  exponents. 

§  44,  pp.  72-73. 
Final  Rule  for  Multiphiing  Monomials:  multijily  coefficients,  add 
exponents.     Exercises  IV.  §  45,  pp.  73-74. 

Division  of  Monomials  :  reverse  multiplication.      Exercises  V. 

§  46,  pp.  74-75. 
Division  of  Poivers  of  the  same  Letter:  subtract  exponents. 

§  47,  p.  76. 

Final  Rule  for  Division  of  Monomials :  divide  coefficients,  subtract 

exponents.    Exercises  VI.  §  48,  pp.  76-77. 

Part  IT.    Multiplication  and  Division  of  Longer  Expressions. 

pp.  78-90. 

Monomial  x  Binomial:  a  (li  +  c)=  ali  +  ac  ;  sum  of  products  term- 
wise.  §  49,  p.  78. 

Monomial  X  Longer  Expression :  sum  of  products  termwise.     Exer- 
cises VII.  §  50,  pp.  78-79. 

Longer  Expressions  -=-  Monomial :  sum  of  quotients  termwise.    Exer- 
cises VIIL  §  51,  pp.  79-81. 

Monomial  Factors:  application  of  rule;  work  by  inspection. 

Product  of  any  Expressions  :  sum  of  jiartial  products.    Exercises  TX. 

§§  52-53,  pp.  82-83. 

Division  of  any  Expressions :  long  division  by  reversal   of  partial 
products;  principally  exact  divisions.     Exercises  X. 

§  54,  pp.  84-88. 


SUMMARY  117 

Polynomial :  letters  occur  in  simple  powers ;  a-\-hx  -\-  cx^  +  •••  +  /x"; 
a  +  ftx  +  cj/  +  (Ix^  +  exy+fy--\ .  §  55,  pp.  88-89. 

Review  Exercises  for  Part  II  of  Chapter  IV:     Exercises  XI. 

pp.  89-90. 

Part  ITT.     Special   Multiplications;    Factors;    Type   Forms, 

pp.  89-102. 

Form   I:    {x -\-  i/Y  ■=  x- -\- 2  xy  +  y'^ ;   polynomial  factors  of   poly- 
nomials only.     Exercises  XII.  §§  56-57,  pp.  91-93. 
Form  II :  (^x  —  y)^  =  x'^  —  2xy  +  y'^.     Exercises  XIII. 

§  58,  pp.  93-94. 
Form  III :  (x  —  y){x  -\-  y)  —  x'^  —  y'\     Exercises  XIV. 

§  59,  pp.  94-95. 
Form  IV:  (x  +  a)(x  +  b)=  x'^  +(a  +  b)x  +  ab.     Exercises  XV. 

§  60,  pp.  95-97. 
Form   V:   ax"^  +  bx  +  c;   factors  by  trial;  cross  products.     Exer- 
cises XVI.  §  61,  pp.  98-99^ 
Other  Forms :  a^  ±  b%  a*  -  b\  («  ±  by,  (a  ±  b)*.     ExerciseTXVII. 

§  62,  pp.  99-100. 
Factoring  by  Grouping :    a(b  +  c)=  ab  +  ac;    review  of  §  51  ;    re- 
arrangements.    Exercises  XVIII.  §  63,  pp.  100-101. 
Review  Exercises  for  Part  III  of  Chapter  IV:   Factoring;  Exer- 
cises XIX.  p.  102. 

I'art  IV.     Applications;    English  translated  into  Algebra. 

pp.  103-115. 

Expression  of  English  in  Formulas  :  search  for  structure  in  examples. 

§  64,  pp.  103-105. 
Simple  Changes  in  Equations :    any  operation  that  gives  a  sinqle 
result  may  be  performed  on  the  lohole  of  each  side. 

§  65,  pp.  105-106. 
If  a  pruiluct  is  zero,  at  least  one  factor  is  zero:  illustrative  «problems 
stated  in  English.  §  66.  pp.  106-109. 

Solution  of  Problems :   complete  check  advisable.     Exercises  XX. 

§  67,  pp.  109-112. 
Review  Exercises  for  Chapter  IV:  Exercises  XXL 

pp.  113-115. 


CHAPTER  V.     FRACTIONS 

PART  I.     COMMON  FACTORS;   REDUCTION  OF  FRACTIONS 

68.  Introduction.  In  dealing  with  fractions  we  shall  use 
the  notation  and  the  rules  of  elementary  arithmetic,  which 
we  shall  review. 

A  fraction  indicates  the  quotient  obtained  by  dividing 
one  number,  called  the  numerator,  by  another  number, 
called  the  denominator.  The  fraction  is  written  by  plac- 
ing the  numerator  over  the  denominator  with  a  horizon- 
tal line  between  them. 
/  The  numerator  and  denominator  are  called  the  terms  of 
the  fraction.  The  form  in  which  a  fraction  is  written  may 
be  changed  in  various  ways  without  changing  the  value  of 
the  fraction,  as  in  elementary  arithmetic. 

OQ     5  OQ     5 . 4 

Thus,  "—  —  -,   for    ^^  = ,  and  the  numerator  and  denominator 

12      3  12      3.4 

may  each  be  divided  by  this  common  factor  4. 

Similarly,  —  =  "  ' '         —  =  — :    here  both   terms  of  the  fraction 
•^    84      2-2  .3-7      14 

are  divided  by  the  common  factor  2  and  also  by  the  common  factor  3. 

The  rule  of  elementary  arithmetic  is: 

"  The  value  of  a  fraction  is  ufichanged  if  both  numerator 
and  denominator  are  divided  hy  the  same  number. ^^ 

We  shall  follow  the  same  rule. 

For  example  :  2n'^b  _    2  •  a  •  a  -b    _  a 

4^2  -  2-2-a-h-b  ~  26' 

where  we  have  divided  both  numerator  and  denominator  by  2,  also 
by  a,  also  by  h,  each  of  wiiich  is  a  common  factor. 

118 


COMMON    FACTORS;    REDUCTION  119 

Instead  of  tlie  above  rule  we  shall  say 

CD     D' 

which  evidently  means  the  same  thing. 

\_C  is  the  cnmmon  factor  of  numerator  and  denominator  of  the 
given  fraction  ;  N  is  the  numerator  of  the  residting  fraction  ;  D  is 
tlie  denominator  of  the  resulting  fraction.] 

This  principle  may  be  proved  by  means  of  the  rules  of  p.  35. 

For  -  =  X,  provided  N^D-  X, 

and  ^li^=  Z,  provided  C  ■  N={C.  D)  ■  X  =  C  ■  {D  .  X). 

But  by  IV,  p.   106,  the   two   conditional  equations  preceded  by 

"provided"  are   equivalent,   for  we   may  divide  both  sides  of  the 

N  C  ■  N 

last  bv  C,  if  C  is  not  zero.     Hence,  —  and  — — -  are  the  same  number 
•'  D  C  •  D 

X,  provided  C  is  not  zero.    This  proof  need  not  be  learned  at  this  time. 
Notice  particularly  that  O  must  not  be  zero. 

Thus,  in  substituting  numbers  in  the  place  of  letters  in  order  to 
cheek  an  example,  or  for  any  other  reason,  if  the  numbers  substituted 
would  make  one  of  these  divisors  zero,  at  any  point  in  the  work,  they 
must  not  be  used.  Again,  if  both  numerator  and  denominator  are 
divided  by  an  expression  that  contains  unJ.nown  letters,  we  should  be 
careful  to  say  that  the  result  is  correct,  prodded  the  (unknown)  value 
of  that  expression  is  not  zero. 

Reversing  the  preceding  rule,  we  may  multiply  both  numerator  and 

denominator   by  the  same  number,  except  zero.     Notice,  however, 

that  adding  the  same  number  to  both  numerator  and  denominator  is 

not  allowable,  in  general.     Thus, 

3                            3  +  2 
-  is  not  equal  to  -; r- 

Like^ase,  subtracting  the  same  number  from  both  terms  is  not  allowable. 

69.  Common  Factors.  In  §  68  we  used  any  common 
factor  of  both  numerator  and  denominator  as  the  divisor 
of  both.  These  common  factors  may  be  found  by  inspec- 
tion in  simple  examples,  as  in  §  51,  p.  79. 


120 
Ex.   1. 
Ex.  2. 
Ex.   3. 


FRACTIONS 

[Cn 

20  cfi¥ 
15  a%^ 

46- 
3a- 

5  aH>^      4  h 
5  a%^     3  a 

x^- 

1 

.  (x-l)(a-  +  l) 

(a;+l)(a;  +  l) 

-1 

4-1 

x^+'2x 

+  1 

m2  +  5 

w  +  4 

_   (m  +  l)(m- 

f4) 

_  m 

+  1 

m^  — 6w  — 40      (w— 10)(m  +  4)      w  —  10 


If  all  the  common  factors  are  removed,  the  fraction  is 
said  to  be  in  its  lowest  terms. 

The  rule  is  the  same  as  in  elementary  arithmetic  : 

To  reduce  a  fraction  to  its  lowest  terms,  divide  both  numer- 
ator and  denominator  by  each  of  their  common  factors ;  in 
the  resulting  fraction  the  numerator  and  the  denominator 
shoidd  have  no  common  factor. 

EXERCISE  I:     CHAPTER  V 

Simplify  the  fractions  : 

■*■■    20"        ■    144*       ■2.33.52'       ■    ^aW'  '     -24a6a;** 

„    65       ,189      ^    4='.5.62      „    XT^Z^      ,_    35mVp 

2     •       4     •      6.    ■ '      8.    .      10.    *-  • 

■    70  105  2.3.4  XY^Z  2lw?np* 

16  mhi'       ^2      -12  2^^         ^3    -lOx'yz^     ^^   ifq-^^ 
24mp2  •    ^2S)xyzt-  '       2y?yh    '        '        p-q- 

21{a  +  h)\x  +  y)h^  ^g     Q>b{q-r){r-x>)(p-ir 

•     15{a  +  b)\x  +  y)r'  '    2Q{q-r)\r-pf{p-l)' 

70.  Highest  Common  Factor.  As  in  elementary  arith- 
metic, the  product  of  all  the  common  factors  is  called  the 
highest  common  factor,  usually  abbreviated  H.  C.  F. 

This  is  often  called  highest  common  divisor,  H.  C.  D. ; 
also,  in  elementary  arithmetic,  greatest  common  divisor, 

a.  c.  D. 


69-70]  COMMON    FACTORS;    REDUCTION  ll'l 

In  arithmetic  we  found  a  number  as  G.  C.  D.  Thus  the  G.  C.  D. 
of  60  and  72  is  12,  for  60  =  2  •  2  •  8  •  5  and  72  =  2  •  2  •  2  •  3  . 3.  The 
common  factors  are  2,  2,  and  3,  and  their  product  is  12. 

Just  here  we  are  working  with  polynomials ;  that  is,  expressions 
each  of  whose  terms  contains  only  simple  powers  of  the  important, 
letters.  (See  §§  55,  56,  pp.  88,  91.)  We  shall  use  as  common  factors  of 
tiro  such  poh/nomials  only  factors  which  are  themselves  polynomials*  The 
H.  C.  F.  here  is  therefore  a  polynomial,  though  we  shall  later  have 
examples  of   another  kind. 

EXERCISES   II:     CHAPTER   V 

Reduce  to  lowest  terms  after  finding  the  H.  C.  F.  of  numera- 
tor and  denominator : 

^    15x-y-10xf  2    12  a%'c  +  18  aba'  + 15  be" 

5  xy^  30  abc 

3  8/)^  g    z'-Bz  +  G^      ^     l-2r-8r^ 

IG pY -  2-i 2>q^'       '     Z-  +  Z-G   '        '    l_7r-18r2' 

^     m^n  —  mn^  ^    z^  —  bz  +  G  x^  —  2  xj/  +  ;f 


nr  —  n^  z-  —  z  —  6  ay^  +  4:  xy  —  5  y'^ 

m^  —  v^  .-6v'^— v  —  2  % 

15. 


m^  +  2  vvhi  —  3n^  9  v-  —  4 

,.2 _ 2  rg -  15  s^  ^Q  3z'-13z  +  12 

'   7^  +  81-8  + 15  s"'  ■  9z--242;  +  16* 

^  •      2u'-uv-v^   '  '  sH-st' 

a;2_2.Tj_35  12x2-a-?/-6v' 

12.     — -; —  •  18. 


2  a;2  -  5  a:  -  63  Sx'-2xy-3  y- 

l_a;_6a^  19    5p'-26pq  +  5q\ 

'    3-x-24:J^'  '    Sp^'-opq-oOq' 

14    t^lt.  20.  a'  +  2ab  +  b'-c' 


x^-  y^  a^  +  b-  +  o^  +  2bc  +  2ca  +  2  ab 

*  It  should  be  noticed  that  an  expression  of  even  a  single  term  may  be 
a  polynomial  ;  it  is  so,  provided  it  contains  only  simple  powers  of  the  im- 
portant letters,  or  if  it  is  a  known  constant.     (See  §  9,  p.  9,  and  §  55,  p.  88.) 


122  FRACTIONS  [Ch.  V 

71.  Practical  Work  in  Common  Factors.  The  chief  use 
of  the  H.  C.  F.  is  in  the  reduction  of  fractions.  In  cases 
where  the  factors  cannot  be  found  by  inspection,  the 
product  of  all  those  common  factors  which  we  can  find  is 
used  in  its  place.  Thus,  practically  speaking,  we  find  the 
H.  C.  F.  only  in  the  simpler  examples. 

Ex.  1.    Find  the  H.  C.  F.  of  36,285  and  44,895. 

We  may  notice  that  5  and  3  are  each  common  factors  (by  trial). 
Hence  15  is  a  common  factor.  The  student  might  try  for  some  time 
without  finding  any  other  common  factor.  We  should  say,  after  a 
reasonable  amount  of  work,  that  15  is  a  common  factor  of  36,285  and 
44,895,  and  that  we  cannot  easily  find  any  others.  However,  41  is 
also  a  common  factor;  the  H.  C.  F.  is  really  15  x  41  =  615. 

There  is  a  long  process  that  is  sometimes  used  to  make  sure  of 
the  H.  C.  F. ;  this  is  hardly  worth  while  at  present,  liowever,  for  the 
work  required  in  it  is  longer  than  the  value  of  the  results  justify. 
The  process  is  found  at  the  end  of  this  book.     (See  Appendix,  §  19.) 

Practically  speaking,  we  wish  to  reduce  a  fraction  only  so  far  as  i? 
convenient.  For  example,  ||ff|  =  |HI  after  taking  out  the  common 
factors  3  and  5  which  are  easy  to  find.  If  we  wish  really  to  divide 
numerator  by  denominator  to  find  the  value  in  decimals,  it  is  worth 
while  to  find  such  easy  common  factors  as  3  and  5  and  remove  them 
first.  But  it  would  be  a  waste  of  time  to  try  past  11  or  13  in  seeking 
common  factors  by  trial.  Certainly  it  would  be  a  waste  of  time  to  try 
a  number  as  high  as  41  —  which  is  really  the  next  common  factor,  for 
the  whole  work  required  to  divide  2419  by  2993  (to  several  places  of 
decimals)  is  less  than  the  work  required  to  find  the  factor  41,  either  by 
trial  or  by  the  long  process  mentioned  above  (Appendix). 

We  shall  say  that  a  fraction  is  reduced  ^.s-  low  as  is  prac- 
ticable, or  is  in  its  lowest  practicable  terms,  when  all  the  com- 
mon factors  of  numerator  and  denominator  that  can  be 
found  by  methods  we  know  at  present  have  been  removed. 

Thus  .  36285  ^  3  •  5  ■  2419  ^  2il9         ^^^^  ^^  practicable), 

44895      3.5.2993      2993  "^ 

although  II  =  g^  =  I   (lowest  terms). 


71]  COMMON   FACTORS;    REDUCTION  123 

Likewise  a"6  +  a^b^  +  a^f>^  _         a^Ha*  +  «'^''  +  b*) 

a(a*  +  a^b^  +  b*) 
~  b-(a^ -  4  a^b  +  4  ab^  -  :j  b^) 

(as  low  as  practicable), 

aithouoh      a(a*  +  a-'b'^  +  b^)  ^a(a^- ah +  b'^){a^ +ab  +  b^-) 

=  -^^ ■ ■ — i  (lowest  terms). 

b\a-:^b)     ^  ^ 

Unless  the  factors  happen  to  be  known,  the  second  operation  for 
finding  the  result  in  lowest  terms  cannot  be  done  without  considerable 
work;  it  is  scarcely  worth  while,  the  result  in  lowest  practicable 
terms  being  the  one  most  often  used.  * 

EXERCISES  III  :  CHAPTER  V 

Reduce  to  the  lowest  terms  practicable  : 

1   2.0  4  6.     o   3315      «3    8346      4,   41710      e   17384 
-^*   2418*     ^*   4845*     "^^      13206'     '•   43430"      **•   18U40" 

g     a?  +  h^  g     6^  +  76'-8  6 

10.     -  -   .  -  ■-  -  3^3     ar^  -  a^y  —  xy-  +  f 

2  m^  +  3  m^in?  —  5  min?  a^  +  x^y  —  xy^  —  y^ 


9636 
38  52 

7. 

mhi 

+  3 

??t-7l^ 

-4 

7n7i^ 

6  ?•  +  5  x^  —  2  x^y  +  2  a;.y^  —  2/^ 

12  ar^.y  -  a- j/3  ^g  3  a^ft^  +  6  a^ft^  +  27  a6« 


2  x^?/  -  5  a;''?/^  +  3  a;y/^  2  a^ft^  _  2  a453  ^  2  a'h''  +  6  a^ft'' 

[Many  of  the  above  exercises  will  of  course  be  reduced  readily  to  the 
actual  lowest  terms.] 

*  The  question  as  to  when  any  fraction  really  is  reduced  as  low  as  is 
practicable  is  to  some  extent  arbitrary,  and  depends  on  the  breadth  of 
the  student's  experience  in  working  with  literal  expressions  and  recogniz- 
ing factors.  If  the  factorization  of  a*  +  a^l>^  +  6*  and  a^  —  4  cfib  +  4  ah^ 
—  3  h^,  used  above,  were  familiar,  then  the  fraction  would  be  in  lowest 

terms  practicable  only  when  written  in  the  form  ^^"   +  <^>+ — i .     ^pjig 

b'^{a—  3  b) 
teacher  will  be  able  to  judge  what  degree  of  reduction  may  be  reasonably 
expected  in  the  case  of  his  own  students.     See  also  Appendix,  §  34,  et  seq. 


PART  II.    RULES  OF  OPERATION 

72.  Rules  of  Signs.  A  fraction  is  merely  an  indicated 
quotient,  as  in  elementary  arithmetic.  The  rules  for 
signs  hold  as  in  division. 

Thus,  . =  —  ;    =  —      and  =  +  - 

b  b       -I,  b  -b  b 

Likewise,    -(^J  =  .,^;     -(^^.f.     -(ff)  =  -r 

There  are  three  important  signs  to  be  considered :  the 
sign  of  the  numerator,  that  of  the  denominator,  that  in 
front  of  the  whole  fraction.  Of  these  three  important  signs 
any  two  may  he  changed  at  the  same  time,  without  changi^ig 
the  value  of  the  fraction. 

The  sign  in  front  of  the  fraction  affects  the  total  result. 

Thus,   in   the    fraction "      '^    ,    if    a  =  8   and   i  =  3,  we   get 

2  a  —  h 

__  3  •  8  +  5  •  3  ^  _  24  +  15  ^  _  39  ^  _  3 
2-8-3  16-3  13 

To  change  the  sign  of  the  numerator  (or  denominator) 
be  careful  to  change  the  sign  of  each  term,  including  the  first 

^^^^  •'  _  3  a  +  5  ^*  ^      -  (3  g  +  5  b)  ^  -3a -56 

2a- b  2a -b  2a -b     ' 

Check :  Let  a  =  8,  ?>  =  3  in  this  answer ; 


-3-8-5-3 
2-8-3 

_  _  24  -  15  _  -  39  _       3 
16-3             13 

Again, 

3  a  +  5  ft 
2a  -ft 

_     3a  +  5ft    _3a  +  5ft_3a  +  56 
-(2o-ft)       -2a  +  ft        ft-2a 

Check : 

Let  a  =  8,  ft  =  3 ; 

3-8  +  5.3_24  +  15  _      39  _       3 
3-2-8         3  -  16        -  13 
124 

RULES   OF   OPERATION  125 

Note,  in  the  last  check,  the   fraction  ~- ;  this  may  also  be 

-  -H  -  15  '^  ~  ^^ 

written  — ,  where  we  have  changed  the  signs  of  both  nuniera- 

-8+16 

tor  and  denominator. 

If  the  numerator  (or  the  denominator)  is  the  product 
of  several  factors,  a  change  of  sign  of  one  of  these  factors 
changes  the  sign  of  the  whole  fraction.     Thus, 

(8  -  x)(x  -  5)  ^  -  (3  -  x)(x-  5)  _(x-  3)(x  -  5) 
12  12  12 

EXERCISES   IV:    CHAPTER  V 

Make  each  of  the  three  possible  simultaneous  double  changes 
of  sign  in  each  of  the  following  fractious ;  check  each  exer- 
cise by  inserting  numerical  values : 

1.     -K  4.     ~^^  7.  5  ^'.V 

h'  —  c^  —y-\-z 

3a-2b  „         k'-P 

—  -iz  +  abc  rs  —  t 

_         abc  —  s^                         t  —  n-\-v  —  w 
o. •  9. • 

p  —  g  u  +  v  —t  —  v+w 

^Q     -(4x-2)(3-5.T)         ^2     (a^b)(b-c) 
5-2x  ■    (c-d)(e-/) 

11.    Simplify ^^.        13.    Simplify  -  (^-'"^X^^-")  . 

^     ^       -(-1)  ^      ^      (4-14)(8-5) 

73.    Addition   and   Subtraction   of    Fractions.       To   add 
several  fractions  we  proceed  as  in  elementary  arithmetic. 
Thus  2    I    4  _  li  4. 11  _  2  2 

^  ""^'  3  ^^  5  —  15  ^  15  —  1^- 

T  .1        .  a  ,    c      a  •  d  ,   c  -h      ad  +  be 

Likewise,  -  +  -  =  _  +  _  =  _^. 

First  reduce  the  given  fraetiovs  to  a  common  denominator  ; 
then  add  the  new  numerators  and  place  this  sum  over  the 
common  denominator. 


2. 


X- 

-y 

y-- 

-z- 

- 

V 

126  FRACTIONS  [Ch.  V 

This  rule  may  be  proved  by  means  of  the  principles  on  p.  35,  as 
follows.     This  proof  need  not  be  learned  at  this  time. 

h  d 

then,  ad  =  h  X  bd  and  cb  =  k  x  bd;    hence,  ad  +  cb  =  (h  +  k)bd  or, 

bd  b      d 

.     «  4  x^     3  w   ^1    ^  •      2  a;v  ,  4  a;2     3  v 

Ex.1.  2xy  +  -^-^^,i\.^i^^,  -^  +  -^-J- 

Taking    the    common    denominator    2  xif,    we    may    write    the 
preceding  expression  in  the  form : 

2  a:.y  X  2  xy"^     4  x^  x  2  a:  _  3  y  x  y"^  ^  4  x^  +  ^j!_  _  ^jL 
lx2xy^         y^x2x       2  x  x  y^     2  xy^       2  xy^     2  xy^ 

4xV  +   8  3.-3-   ^yi 

2  xy'^ 

-c^        o  ,11^  —  1         r,       X  +  1 

Ex.2.    x-\-l-] -  —  Z-  - 

X+1  X— I 

(x  +  l).(x^-l)      (x-l)-(x-l)      2(3:  +  1)  •  (x  +  1) 

1.(X2-1)  ^(X  +  1).(X-1)  (X  +  1)-(X-1) 

(j:S  +  x^-  X  -  l)  +  (a:2-2:r  +  l)-2(x-^  +  2a;  +  l) 

X2-   1 

_a;8  +  a;2_3._i  +  a;2_2x  +  l-2j:2-43:-2^  x^-1  x~2 

-  ^2^^  ^2-1 

In  these  examples  one  of  the  fractions  has  the  sign  — . 
This  amounts  to  subtracting  it  from  the  others ;  in  any 
case  the  result  is  called  the  algebraic  sum.  As  in  ex- 
ample 1,  if  a  term  is  inserted  to  be  added  to  the  others 
which  itself  is  not  written  as  a  fraction  (2.xy  in  Ex.  1), 
such    a   term    may  be  written  in  the  form  of  a  fraction 

(:L^^  for  example)  by  inserting  the  denominator  1. 

Such  a  form  as  2  xy  +  ^-^  +  '^-^  is  often  called  a  mixed  expression, 
_y2        2  X 

since  part  is  in  fractional  form  and  part  is  not. 


f3-74]  RULES   OF   orERATlON  127 

As  common  denominator,  any  convenient  expression 
may  be  taken  of  which  each  given  denominator  is  a 
factor ;  in  any  case  the  product  of  all  the  given  denomi- 
nators will  surely  suffice.     See  also  §  74,  below. 

EXERCISES    V  :   CHAPTER   V 
-,    a;-l      2x  +  3  ^   y-z  ^  2z-5x  ^  x-y 


3  2  ^x  1 X  X 

2  ^    ^    >  4  X         x+1 

11  ^11 

ah  a     0 

■      1    +-J-.  e.     1  1 


a;  +  aa;  — a  x  +  a      x  —  a 

g_  x  +  2  ^  x  +  l      2(x-''  +  3a;  +  l). 
'  x-\-3         X  x^  +  3x 

[The  additions  worked  out  in  Exs.  3,  4,  7,  8  are  very  often 
useful,  and  are  worth  committing  to  memory.] 

,0    -^ -^^±1 ^.  12.    t^--^^-l. 

x-i-2     x  —  2     X-  —  4:  x-\-y     y  +  z 

11.   1 1__^.  13.  _^4- J_+     1 


w      l-^-io      IV-  y  —  zz  —  xx  —  y 

14.  3-      ^x  +  y-\-z  2y{x  +  y  +  z)  ^  ^ 
'y  +  z         y-z  2/'-2!^ 

15.  -^-^+  '-" 


5_^     2-x     10-7a;  +  .x2 

74.  Common  Multiples.  In  adding  fractions  above,  we 
have  used  as  common  denominator  any  convenient  ex- 
pression of  which  each  given  denominator  is  a  factor. 
Such  an  expression  is  often  called  a  common  multiple  of 
the  given  denominators.  In  general,  a  multiple  of  any 
expression,  just  as  in  arithmetic,  is  an  expression  of  which 
the  given  expression  is  a  factor,  or,  in  other  words,  an 


128  FRACTIONS  [Ch.  V 

"exact"  divisor.  We  refer  here  to  polynomials,  and  we 
seek  a  common  multiple  of  which  each  of  the  given  poly- 
nomials are  factors  (see  §§  55,  70,  pp.  88,  121) ;  similarly 
in  arithmetic,  if  integers  are  given,  we  seek  a  common 
multiple  that  is  an  integer. 

The  product  of  several  expressions  is  evidently  a  com- 
mon multiple  of  them ;  for  each  of  them  is  a  factor  of 
their  product. 

Often  some  simpler  common  multiple  can  be  found. 

Thus,  given  72  =  2  •  2  •  2  •  3  •  3,  and  39  =  3  •  13,  and  48  =  2  •  2  •  2  •  2  •  3, 
one  common  multiple  is  the  product  of  72  x  39  x  48.  A  simpler  one 
is  found  by  omitting  in  this  product  all  repeated  factors  that  occur 
in  two  or  more  of  the  numbers.  Thus,  (2  •  2  •  2  •  3  •  3)  x  (13)  x  (2)  is 
a  common  multiple  ;  indeed  is  the  least  common  multiple,  L.  C.  M.,  of 
the  given  numbers,  i.e.  it  is  the  least  number  of  which  each  given 
number  is  a  factor. 

Likewise,  given  a^  —  P,  4  (a  +  ^),  and  (a  —  by,  one  com- 
mon multiple  is  their  product: 

But  a'^-b'^  =  (a-  h){a  +  6),  (a  -  by  =  (a  -  6)(a  -  6),  and 
4(a-f  6)=4(a  +  A). 

Hence,  a  common  multiple  is 

[(a-6)(a  +  ^')][4][«-^], 

where,  just  as  before,  we  have  omitted  duplicate  factors. 
This  is  the  lowest  common  multiple*  (L.C. M.),  i.e.  if  avy 
of  the  factors  now  remaining  in.  the  result  tvere  omitted,  the 
result  would  not  be  a  common  multiple. 

The  rule  is  precisely  like  that  of  elementary  arithmetic  : 
Factor  each  of  the  given  expressions;  write  as  their  L.  CM. 
the  product  formed  by  omitting  in  each  any  of  its  factors 

*  It  is  usual  in  arithmetic  to  use  the  word  "  least,"  in  algebra  the  word 
"lowest,"  in  this  connecliou. 


74]  RULES   OF   OPKRATION  lliU 

that  is  already  written  down  as  many  times  as  it  occurs  in 
the  expression  under  consideration  ;  in  short,  omit  the  dupli- 
cate factors.  Care  must  be  taken  to  have  a  factor  occur 
in  the  L.C-M.  as  many  times  as  it  appears  in  any  one  factor. 
Practically  it  may  not  be  convenient  to  factor  the  given 
expressions  completely ;  in  this  case  we  omit  only  tliose 
duplicate  factors  which  can  be  found  by  the  methods 
known  at  present.  The  lowest  practicable  multiple  is  thus 
found ;  it  is  the  simplest  common  multiple  that  can  be 
found  conveniently. 

Thus,  given  44,895  and  36,285,  we  find, 

44,895  =  3  •  5  .  2993. 
36,295  =  3.5. 2419. 

Hence,  the  lowest  practical  multiple  is 

(3  •  5  •  2993)  X  (2419)  =  108,601,005. 

As  a  matter  of  fact  (see  p.  122)  41  is  a  factor  of  2993  and  also  of 
2419.     Hence  the  L.  C.  M.  is 

(3  .  5  .  41  •  73)  X  (59)  =  2,648,803. 

In  this  example  it  is  clear  that 

T    n  A/T        Product  of  the  two  given  expressions 

^■^■^'-  hTcIf:  ■ 

This  is  true  if  only  two  numbers  are  given,  for  the  H.  C.  F. 
is  exactly  the  product  of  those  duplicate  factors  that 
are  to  be  omitted  in  forming  the  L.C.M. 

This  remark  leads  (see  Appendix,  §  19)  to  a  general  method  for 
finding  L.C.M.  in  any  case;  but  this  process  is  usually  long  and  is 
scarcely  justified  by  the  value  of  the  result. 

In  adding  fractions  by  the  rule  of  §  73,  p.  125,  choose  as 
the  common  denominator  the  lowest  practical  multiple  of  the 
given  denominators ;  then  reduce  each  fraction  to  this 
denominator  by  multiplying  its  numerator  and  its  denomi- 
nator by  the  quotient  formed  by  this  common  denominator 
divided  by  the  given  denominator,  and  proceed  as  on  p.  125. 
hedrick's  el.  alg.  —  9 


130  FRACTIONS  [Ch.  V 

Ex.1.      2a;i/       2x-y      x^  +  2y\ 
^  —  y-       x-y        (.^•  +  2/f 

We  notice  that  the  L.  C.  M.  of  the  denominators  is 

(x  -  y)  {x  +  y)  {x  +  y)  =  (x  -z/)  (x  +  y^. 

Hence,  the  expression  given 

_      2  xy{x  +  y)      _  (2  a:  —  ?/)(x  +  y)^  _  (x^  +  2?/^)(x  —  y) 
(x  -  y) (x  +  ?/)2       (x  -y){x+  y)'^  (x  -  y)(x  +  yY 

_  (2  x^y  +  2  x?/2)  _  (2  xs  +  3  x^y  -  y3)  -  (x^  -  x^y  +  2  xy^  -  2  y^) 
(x  -y)(x  +  yy 

_    -  3  x3  +  3  y3    _  _      3  (x«  -  y^) 
(x  -  y)  (x  +  yy~      (x  -  ?/)(x  +  yf 

_  _  3  (x  -  ?/)(x^  +  xy  +  y^)  _      3  (x'-^  +  xy  +  y^) 
(X  +  y)(x  +  y)  (x  -  y)  (x  +  y)^ 

In  tliis  example  the  result  is  reduced  to  its  lowest 
terms  ;  the  student  should  try  to  do  this  in  every  example, 
and  the  result  should  be  reduced  to  lowest  terms  or  at  least  as 
far  as  is  practicable.  The  student  is  advised  to  leave  the 
common  denominator  in  factored  form,  as  in  the  preceding 
example,  until  the  work  is  completed. 

EXERCISES   VI:  CHAPTER   V 
Perform  the  indicated  operations  : 

11  „  1 


+ 


a  +  rt"     a  —  a^  a^  +  2  x  —  15     x-  —  x  —  Q 

X x  —  y  4  x  —  1 X  —  4: 

x^  +  xy     x~+2xy-\-y'^        '    a;2  +  2j;  — 15     ar  —  x  —  G 

g      x  +  y L.  _A &i±lyi 


y  {y  +  ^)    y{y-  ^)    y"  -  ^ 


71-7o]  RULES   OF   UPEllATlUN  131 

7.  -. — ^— +  ^-i — .+       ' 


/\. 


{x  -  u)  {x  -z)      {y-  z)  (y  -  x)      (z  -x){z-  y) 

2r-l  r-2 1 

,•2  +  2  r  — 3     ?-^  — 1      r  +  3 

x—\  x—\  4 


0^  —  5  x  +  6      ar  —  a;  —  2      a;-  —  2  x  —  3 


10.    ^-i +         1 


X-  —  3  a;  —  4     a;-  —  5  a;  +  4 
11      _2^L_  I      1  1  4^'" 


.r  +  /      x-  +  ^      y  —  x      x*  —  y* 
12.     ^:»^_1^-+  7 


13. 


2  a; +  3      2-a;     2a---a;-6 
2  3 


if2  _  2  .c  +  1      ar'  -  3  a;^  +  3  x  -  1 


14  ^        I        1       _     <"(•»-!)     . 

3.c  +  21-2a;     6.c-  +  a;-2 


v/l5.  ^'  +  ^ 2^ 


+ 


2a;2  +  5a;  +  2      42:^  +  5  a;- 6      8  ar'- 2a;-3 

75.  Multiplication.  We  have  already  found  and  stated 
the  rules  for  multiplying  fractions  (see  p.  67)  ;  these  are 

a       ,  _tto  ^    a      c  ^ac 
h  h  ^    h      d.     hd 

In  words  these  rules  read  as  in  elementary  arithmetic : 

TJie  product  of  two  or  more  fractions  is  the  product  of  their 
numerators  divided  hy  the  piroduct  of  their  denominators. 

Notice  that  this  rule  may  be  applied  when  one  of  the 
given  factors  is  not  in  fractional  form  if  we  supply  such 
a  factor  with  a  denominator  1. 


132  FRACTIONS  [Ch.  V 

A  factor  that  appears  in  the  numerator  of  any  of  the 
given  factors  may  be  canceled  with  the  same  factor  in 
the  denominator  of  any  other  one,  for  this  amounts  to 
dividing  numerator  and  denominator  of  the  result  by  the 
same  factor. 

Thus, 

'    ■    35     14     ^  .  7      ^  .  7      7  .  7     49  ^ 

Ex  •>      ^g''  ^96^^     4  V     ^-S.b.p^^Sb  _ 
■"■    33 6«     20a'     ^  -  11  •  j/     ^  -  5  •  a  ■  r/     55a 

Ex  3         /^-ly./a^-iy      (x-l)(x  +  l)      x-1 
■    *        \3x  +  2j     [x  +  lj  3r»  +  2  x  +  1 

^      (x-iy 
■      3x+2* 

Do  not  fail  to  take  account  of  the  rule  of  signs  in 
multiplying. 

The  result  should  be  reduced  to  lowest  terms  or  as  nearly 
to  lowest  terms  as  is  practicable. 

EXERCISES  VII  :  CHAPTER  V 

Perform  the  following  multiplications: 

1  fiQ        —00 

1        16y49  2        30    y    22  3        __i±^     y  ^'^  . 

■*■'      21  ^   32'  **•     T7   ■^  T5-  "'•  _  72  91 

^^     X--^X=|^.  8.    1x4x1x4. 


-  35  -  22        27 


3    ^    D   ^   7 


15  x'y     16  yz  35  b'c'      24  a'b' 

'      8z'    ^  4:5x'  '     -12a-^  -25c^' 

u  —  V            V  +  u  -^     a     b      c      d 

u^  -\-2uv  +v^      v  —  u  b      c     a     a 

„     l  —  t-20f,      1  —  t  ,_     p-q   ,  f    ■,        2\ 

7. •  11.    ^ X  (  »"  —  0"*). 

1-t'        ^l-Wf  p  +  q       ^^       ^^ 


75]  RULES   (JF    Ol^KRATION  188 

x^-5x  +  G     or  — 7  X +  12 
•    a^-5x+4:^  a^-lx+lO' 

13         «-^      X   a-  +  2  a  +  1 


a-  -  a  -  2      cr  -  2  ab  +  b^ 


14      -l''>(a---«)-(y-^)'''  ^  77  (x-a){z-c) 
14  (2 -c)-  -25  (2/ -6/ 


15.    ! — .  X 


ax  —  ar      ar  +  3  ax  +  2  a- 

16.  a.-3-y?  ^a;-^4-4a;y  +  3y^ 
.r  +  7  .r?/  +  6  y-        x-  +  xy  +  y'^ 

17.  a;y  +  y-  ^  j?a;  - jjy  -qx  +  ^y 
p  -  ^  a^  +  y* 

^3    a'  +  b-  +  c?  +  2bc-2ca-2ah  ^a?-h''  +  2bc~c\ 
a+b  —  c  a^  —  b-  —  2bc  —  c^ 

19.       a;^  +  ary  +  a-.y-  +  f     ^^  —  f 
ar*  —  3  a^-^y  +  3  xy'—  y^      af*  +  y* 

20     4a^  +  .r-3     3af^  +  5a;-2 
a.-2  +  a;-2       8a^-2a;-3" 


«,      6  a-— a  —  2        a-  —  1 
21.    X 


a^-a-2       4  a-'  -  1 


22,        ^y  +  ^y      X  ^  ^'  ~  ^^  ~  y^ 
Qx^  +  xy  —  y-  2i?y^ 

23  ""'-y'         X       -'^^  +  2y' 

12  .T^  -  a-2y2  _  6  y^      x^-x-y-xf  +  f' 

24  a«  4-  2  a^6  +  2  a&^  +  b^      a^ -\- S  a%  +  ^  al^  +  ^ 
a^-Sa^b  +  Sab'-Sb'^  a^-b^ 


134  FRACTIONS  [Ch.  V 

76.    Division  :  Complex  Fractions.     We  have  also  found 

and  stated  (see  p.  74)  the  rule  for  dividing  one  fraction 

by  another  ;  it  is  , 

"^  a  ^   e  _ad 

h      d      he 

The  proof  consists  in  the  fact  that  the  quotient  x  the  divisor  — 

the  dividend  : 

ad      c  _a 

he     d     b 

In  words  this  rule  reads  as  in  elementary  arithmetic: 

To  divide  one  fraction  hy  miother,  multiply  the  dividend 
by  the  divisor  inverted. 

r>  a       c      ad      a      d 

tor  -  ^  -  =  —  =  -x-. 

b      d       be      b      c 


T.      ^       x^-l       x  +  1  af-1       x-1      (x-iy 

3x  +  2      x-1  •3x  +  2      x-\-l      3a;  +  2 

Ex.  2. 

^■^ - 1  ^ a;  +1  =  ^~^  ^ a;  +  l  ^  x-— 1  ^     1     _ 


•Sx-\-2  3  a; +  2         1         3x  +  2      a;  +  l      3a;  +  2 

In  this  example  the  divisor  is  a  fraction  only  after  we  insert  the 
denominator  1.  Though  we  should  not  usually  insert  this  denomi- 
nator, it  is  often  convenient  to  do  so. 

The  result  of  inverting  a  fraction  is  called  its  recipro- 
cal.    Thus,  the    reciprocal    of    -    is   -.     In   general,  the 

0  a 

reciprocal  of  any  number  is  1  divided  by  that  number. 
The  reciprocal  of 

4  IS-;  of-isl-^-  =  4;  of-islH--  =  lx=-- 

4  4  4  6  0  a      a        \ 

The  division  of  one  fraction  by  another  is  often  indicated 
by  means  of  the  fraction  form.  In  this  case  the  whole  is 
called  a  complex  fraction. 


r«]  RULES   OF   OPEKATION  135 

a;-  +  l 

ar-l  ^  x^  +  l  ^  x  +  1  ^  ar  +  1      x-1  ^    a^  +  1 

a;  +  l      jr^-1  '  x'-l      ar'-l      a;  + 1       {x  +  iy' 

x-1 

-,         a         a  +  b         a  b 


Ex.  2. 


a  +  ^      a  +  b      a  +  b      a  +  b 


1+-^      ^^^  + 


a  —  b      a  —  b      a  —  b      a  —  b 


b         a  —  b_b(a  —  b) 


a  +  b      a  —  b      a  +  b         a         a  (a  +  b) 

If  there  are  more  tlian  two  horizontals,  great  care  should  be  taken 
to  mark  the  main  horizontal  line  heavily,  for  neglect  to  do  so  may 
lead  to  serious  error.     Thus,  q 

2       ,      5      10       ,.,32^1       2 
_  =  o  V  -  =  — ,  while   _  =  -  X  -  =  —  • 
33       3  535      15 

5 

Similarly,  the  position  of  the  main  horizontal  line  is  essential  in 
fractions  in  algebra,  as  in  arithmetic;  a  mistake  in  recognizing  which 
is  the  main  horizontal  line  causes  a  mistake  in  the  result. 

EXERCISES   VIII:    CHAPTER   V 

What  is  the  reciprocal  of : 

■   "  '    2"         ^''         2' 
2.   a?    ^?    -a?    --? 


a 

a 

b        b         -b 

b' 

Perform  the  following  divisions : 

4    15  .       5^ 
■   16  ■       4 

-64  .         40 
■    -65  ■       -39 

8. 

—  H-  100. 

76 

5         2^  •       ^^ 
21  ■       14' 

144  .  126 
■    25    ■  125 

9. 

4    .  -1 
25  ■  100 

136  FRACTIONS^ 

-  12  d'b^c  .    2AaH)  ^^  50  rV  .  75afy 

°'       35a?y      '  -Ixy'  '  49^^    '    562=^  * 

36  hc^     32  ah  ^^      4m^     .  - 12  ?m 

25a    ■     15    *  ■  -9/71^  •        n^ 

-  65  j-^  .  143  r.s'^ 

■      64s2  ■  -144' 


16  a;y  ^  a^^  +  2  x?/  _ 

x'  +  2xy  +  y-  '     x^-y- 

-2ab-3b'       a'-b^ 


17. 


a^-2ab  +  b-    '  a-b-ab^ 


18     p'  +  pg-^q'  .  p'-3pq  +  2q\ 

p^  —  q-        '        p^+pq 


z^  +  82  +  15  .  z'  +  3z-10 
■    2^_6z  +  8    ■    ^2-5^  +  4  ' 

z^-16      2-z-z' 


20. 


^-1      '1+2  +  2- 


Simplify  the  complex  fractions : 

6  x"  +  a;  -  12  ?/-  -  25 

4  +  4                 ^^     2.'r  +  a;-6  r  +  7i/  +  12 

21.    „        .         .   •  23.        ^    n TTT-  •         25 


1  +  1  + J^'  90.-^-16  ■   /  +  10?/  +  25 

a;-'  +  4  a;  +  4  y-  —  y— 12 

si^  +  x  —  6 


x'-2x-3  1  ^:^  +  ^^dll 

^^    a^  +  2  X-  -  3 '  x-^-5a;  +  6         ^g     /«+l_Jilll 


ar'-So;  1  '  a:  +  l 


2ar-9.7:  +  10  a;-l      ^  ^ 


PART   III.     PROPORTION 

77.    Definitions   and  Introduction.      If  two  fractions  are 
equal,  their  four  terms  are  said  to  be  in  proportion. 

Ex  1.    f  =  I,  for  each  is  equal  to  |. 

This  is  sometimes  written  8:6  =  4:8,  and  read  "3  is  to  6  as  4  is  to 
8,"  but  there  is  no  advantage  in  doing  so. 

Ex  2      ^"^-^    =  ^'  +  1  ^^^^.  ^+1  ^ a;  +  l  ^x  —  l_  x' -  1 


{x  —  iy     x  —  l         x—l      x  —  l      x—1      (x  —  1)- 

Likewise  all  eciualities  of  two  fractions  given  above  are  propor- 
tions. Any  equation  is  a  proportion  if  the  denominator  1  is  supplied 
in  case  no  denominator  is  given. 

A  fraction  is  often  called  a  ratio,  and  -  is  written  in  the 
form  a  :  b,  and  read  "  the  ratio  of  a  to  i." 

78.  Standard  Changes.  We  may  make  a  new  propor- 
tion (equation)  from  an  old  one  by  the  operations  of 
§  65,  p.  106.  Some  of  the  forms  derived  have  been  given 
names. 

I.  If  -  =  -,  we  may  add  1  to  each  side  ;   then, 

h      d 

a  ,  -,       c   ,\  a  +  h      c  -^  d 

h  d  h  d 

This  is  called  the  process  of  composition. 

^     1     2     4,  2  +  3     4  +  6   _   5     10 

Ex.1.    _=-,hence,  -3-=-^,or,-  =  -. 

Ex  '>     ^--.'/^        ^'-f       ■  hence    -1^  =    2^  +  2x;,   , 
■"■    x+?/     x--\-2xy  +  y''  'x  +  y     x'  +  2xy+y' 

II.  If  -  =  -,  ive  may  subtract  \  from  each  side;  then, 

b      d 

a      -,       c      -,  a  —  b      e  —  d 

1= 1,  or    — , —  =  — -— • 

b  d  b  d 

137 


138  FRACTIONS  [Ch.  V 

This  is  often  called  the  process  of  division,  but  this  name 
is  not  well  chosen,  since  ''division"  means  something  else. 

TTT     Ti-  «      <?    i-i         a  +  h      e  +  d        -,    a  —  b      c  —  d 
III.    If  -=-,  then    -^  =  — ^   and   --—^^ — —  • 
b      d  b  d  b  d 

Dividing  the  right-hand  sides,  and  also  the  left-hand 

sides,  we  find, 

a  +  b      c  +  d 

b  d  a  +  b      e  +  d 

or 


a—b      c  —  d  a  —  b      c—d 

b  d 

This  is  called  the  process  of  composition  and  division. 

IV.  If  -  =  -,  ^ve  may  divide  1  by  each  side,  if  neither  is 

b      d 

11  b      d 

zero;  then  —  =  -,     or,     -  =  — • 

a      c  a      e 

b     d 
This  is  called  the  process  of  inversion. 

V.  If  -  =  -,  we  may  muUijdy  each  side  by^  bd;  then 

b      d 

-  X  bd  =  -  X  bd,  or,  ad  =  be. 
b  d 

This  is  called  the  process  of  clearing  of  fractions  or  the 
process  of  cross-multiplication. 

VI.  Finally,  if  ad  =  be,  we  may  divide  each  side  by  bd, 
if  bd  is  not  zero,  [or  l)y  dc;  or  by  ab  ;   or  by  ac],  then 


[ 


ad     be  /i\  ^      <^  . 

n=Tl'^  or,     (1)-=-; 
bd      bd  b      d 

^ox  ^       b  /-ON    d      c  r  i\  ^      ^ 

or,  (2)  -  =  -  ;   or,  (3)  -  =  -  ;    or,  (4)  -  =  - 

e      d  b      a  c      a_ 


78-79]  P1U)1'()RTI(3N  139 

A  comparison  of  these  equivalent  forms  shows  several 
permissible  changes  that  are  sometimes  given  names  ; 
thus,  the  change  from  form  (1)  to  form  (2)  is  called 
alternation  ;  the  change  from  form  (1)  to  form  (4)  is 
called  inversion ;  but  the  simplest  way  to  remember  all 
of  these  is  to  remember  tlie  operation  of  clearing  of  frac- 
tions, No.  V  above,  and  the  principle  stated  in  VI,  which 
amounts  to  dividing  both  sides  by  the  same  number. 

Besides  these  six,  many  other  changes  are  allowable ; 
for  example,  we  might  add  2  to  each  side,  or  subtract  3 
from  each  side,  etc.  In  general  we  merely  carry  out 
operations  that  are  allowable  with  any  equation.  These 
principles  are  often  valuable  in  solving  equations. 

EXERCISES  IX  :  CHAPTER  V 

In  the  following  exercises,  apply  to  each  given  proportion 
the  process  referred  to  by  Roinan  numerals : 

x-2     IG'  3         1  X-       8 

4-    "§^^1^^;=.^;  I,  II,  III,  IV.      5.    |  =  ^;V,VI(3). 

6.    ^^-ti  =  ^JZl/ ;  I,  II,  III,  IV.     7.    ?^^  =  A  .  I,  II,  III. 
a  —  b      x-\-y  ar  +  4      13 

=  ?;   I,  II,  III,  IV,  V. 


2x~  A      3  m"  -\-  rr     inr  +  ^ 

I,  II,  VI  (2). 

79.  Variables  in  Proportion.  Proportions  (equalities  of 
fractions)  arise  whenever  two  varying  quantities  are  so 
related  that  their  quotient  is  always  the  same.  (See  §  20, 
p.  25.) 


140  FRACTIONS  [Cm.  V 

Thus,  if  butter  is  30  cents  per  pound,  n  pounds  cost  30  n  cents,  or 

j9  =  30«, 
where  j9  is  the  cost  in  cents  and  n  is  the  number  of  pounds.    (See  p.  23.) 
If  n  =  3,  JO  =  90 ;  if  n  =  4,  p  =  120 ;  etc. ;  and  we  have 

90      120        ,  «     ,  .3^    , 

—  =  —  =  etc.  =  i-  always  =  30  always. 
3        4  n 

In  case  two  varying  quantities  y  and  x  have  a  constant 

y 
quotient,  i.e.  -  —  k,  where  k  is  constant,  any  two  pairs  of 

values  of  i/  and  x  form  a  proportion,  and  we  say  that  the 
variable  quantity  ?/  is  proportional  to  the  variable  quan- 
tity x;  that  is,  any  pair  of  values  of  y  and  x  form  a  propor- 
tion with  any  other  pair.  We  have  already  used  such 
proportional  quantities  and  have  drawn  corresponding 
figures,  which  we  found  to  be  straight  lines.  (See  pp. 
20-25.)     We  shall  now  make  clear  that  the  figure  is  always 

a  straight  line  if  ^  =  ^,  i.e.  if  y  =  k  •  x. 

X 

80.    Graph.     In  dealing  witli  an  equation  of  the  form 

y  =  kx, 

we  found  always  a  straight-line  figure.  (See  p.  25.)  We 
see  that  if  ^_;l^     ^_^. 

a;  =  2,     y=^k; 
a;=  3,     y=%k; 
etc., 

and  we  wish  to  show  that  these  points  lie  on  a  straight 
line.  To  plot  the  point  x—\yy  =  k\\e  go  one  unit  to 
the  right  and  k  units  upwards  to  the  point  marked  A. 

To  plot  the  point  3;=  2,  y  =  2k  we  go  one  unit  beyond 
A  to  the  right  and  k  units  above  A  (Fig.  20). 

The  rectangle  whose  corners  are  0  and  A  is  precisely 
the  same  shape  and  size  as  the  rectangle  wliose  corners  1 
are    A  and  B.      Hence,  the    diagonal    OA    has    the    same  I 


70-.S()J 


PKoroirnoN 


Ul 


direction  as  the  diagonal  AB ;  consequently  AB  is  an 
extension  of  the  straight  line    OA. 

The  argument  is  the  same  from  B  on  to  the  next  point 
C;  from  0  on  to  I),  etc.  Likewise  backward  from  0  to 
L,  thence  to  M,  etc. 

If  smaller  steps  are  taken,  the  argument  is  always  the 
same.  Thus,  steps  half  as  wide  and  half  as  high  would 
again  bring  us  to  the  same  line. 


M 


N 


/ 


12       3      4 


Fig.  20. 


The  idea  of  steps  —  of  actual  stair  steps  —  is  a  good  illustration  of 
this;  the  broken  line  one  unit  to  the  right  and  k  up  makes  such  a 
stairway.  The  edges  of  these  steps  make  a  straight  line  as  on  any 
ordinary  stairway. 

Finally  we  may  say,  The  equation 

y  =  kx 

is  represented  hy  a  straiyht  line  through  the  starting  pointy 
which  rises  k  ivnlts  in  passing  1  u^iit  to  the  right. 


142 


FRACTIONS 


[Cii.  V 


Ex.  1.    In  the  problem  just  mentioned 

^)  =  30  n, 

where  p  is  the  price  in  cents  of  n  pounds  of  butter.  We  may 
now  draw  the  corresponding  figure  much  more  easily  than  on 
p.  24  where  the  same  example  was  given. 

If  we  draw  only  two  points  of  the  figure  and  coiniect  these  by  a 
straight  line,  we  know  the  figure  is  correct,  since  we  know  in  advance 

that  it  is  a  straight  Hue.  If  n  =  0, 
J)  —  {)  (that  is,  no  pounds  costs  no 
money)  ;  if  n  =  \,  p  =  30 ;  if  ?i  =  5, 
p  =  150. 

Plot  the  point  n  =  0,  p^O  at  A. 
Plot  the  point  n  —  5,  p  —  150 
at/?. 

Join  A  and  B  by  a  straight  line. 
This  straight  line  is  the  desired 
figure ;  from  it  can  be  read  oif  at 
once  the  answers  (approximately) 
for  a  variety  of  problems  :  the  cost 
of  a  given  number  of  pounds  ;  the 
number  of  pounds  that  can  be 
bought  for  a  given  amount  of 
money,  etc.,  as  on  p.  20. 

Thus,  having  worked  the  two 
simplest  examples  of  which  we 
can  think  (that  is,  if  n  =  0,  ;?  =  0 
and  if  n  =  1,  p  =  30),  we  have  in  this  figui-e  the  solutions  (approxi- 
mately) for  any  problems  in  proportion  which  might  be  given  under 
this  example. 

It  is  well  to  plot  a  third  point.  Thus,  if  n  =  10,  p  =  300.  Plot  this 
point  at  C.  It  lies  on  the  straight  line.  If  it  did  not,  we  shoidd  knoiv 
that  we  had  made  some  error.  The  value  in  plotting  three  points  lies 
in  this  check  on  the  accuracy  of  the  work. 

Ex.  2.  In  the  Fahrenheit  and  Centigrade  thermometers  the 
scales  of  temperature  are  made  in  divisions  that  are  propor- 
tioned. Thus  9  divisions  on  the  Fahrenheit  scale  are  equiva- 
lent to  5  divisions  on  the  Centigrade  scale.  If  the  temperature 
rises  9  degrees  by  a  Fahrenheit  thermometer,  it  rises  5  degrees 


Fig.  21. 


SO] 


rKoroR'j'ioN 


143 


by  a  Centigrade  thermometer.  The  amount  of  rise  or  fall  on 
either  scale  can  be  found  by  ordinary  proportion,  in  case  the 
rise  or  fall  is  known  on  the  other. 

But  the  Fahrenheit  scale  is  marked  32°  at  the  freezing  point, 
whereas  the  Centigrade  is  marked  0°  at  the  freezing  point. 
We  must  therefore  subtract  32  from  each  Fahrenheit  tempera- 
ture before  we  can  conveniently  compare  it  with  Centigrade 
temperature. 

Hence, F— 32  is  the  quantity  to  be  compared  with  C,  if  F 
and  C  stand  for  the  readings  of  the  two  thermometers  at  the 
same  time  and  place.  Since  these  compare  as  9  compares  to 
5,  we  have 


F-32 

G 


9 
5' 


and  we  say  that  F—  32  is  proportional  to  C,  the  constant  quo- 
tient (or  ratio)  being  1. 

Multiplying  both  sides  by  5,  also  by  C,  we  have 

or,  5  F  -  160  =  9  C. 

To  draw  the  corresponding  picture,  we  may,  as  on  p.  23,  take  this 
equation  (which  is  given  in  the  Tables  without  argument)  and  plot 
several  i^oints.    We  need  only  plot 
two   of   them,   though   three   are 
better. 

If  C  =  0,  F=32,  as  above  (this 
is  the  freezing  point). 

If  C  =  20,  F  =  68  (this  is  the 
ordinary  temperature). 

If  C  =  100,  F  =  212  (this  is  the 
boiling  point  of  water).  Plot 
tliese  points  ;  they  are  L,  M,  N, 
respectively,  in  the  figure ;  the 
straight  line  through  these  is  the 
graph. 

Notice  that  it  is  here  not  F 
but  rather  F  —  32  which  is  pro- 
portional to  C.  Hence,  the  dotted 
line  which  is  just  32  points  below  Fig.  22. 


144  FRACTIONS  [Ch.  V 

the  line  LMN  is  the  line  which  represents  the  proportion  of  i*'  —  32 
anil  C.  The  heavy  line  really  represents  the  corresponding  values 
of  F  and  C. 

Given  values  of  F,  we  can  find  the  corresponding  C  either  from 
the  figure  (approximately)  or  from  the  equation 

5  F  -  160  =  9  C, 
and  conversely  F  can  be  found  if  C  is  given. 

Thus, if  a  Centigrade  thermometer  reads  28"^,  find  the  temperature 
in  the  ordinary  (Fahrenheit)  scale.  In  the  figure  we  move  out  along 
the  line  marked  C  to  28,  then  directly  upwards  till  we  meet  the 
straight  line  LMN.  The  height  is  the  value  of  F\  it  is  seen  to  be 
82,  approximately.     From  the  equation, 

5  F  -  160  =  9  C  =  9  X  28  =  252. 
Add  160  to  each  side  :      5  F  =  2.52  +  160  =  412. 
Divide  each  side  by  5  :        F  =  82.4. 

The  exact  answer  is  therefore  82°.4;  notice  that  the  answer,  82°, 
obtained  from  the  figure,  is  not  exactly  correct. 

In  any  case  it  is  well  to  get  a  result  from  the  figure  and  a  result 
from  the  equation ;  the  result  from  the  figure  is  only  approximately 
correct;  that  from  the  equation  is  exact.  The  result  from  the  figure 
serves  as  a  check  on  the  exact  answer  from  the  equation. 

In  many  cases  an  exact  answer  is  not  necessary.  In  the  tempera- 
ture example  above,  it  is  often  desirable  to  know  approximately  the 
temperature;  in  fact, very  few  thermometers,  except  those  especially 
made  for  experiments  in  physics,  etc.,  are  really  accurate.  The  error 
made  above  (/^  of  a  degree)  is  smaller  than  the  error  made  by  most 
thermometers  in  ordinary  use.  The  student  should  in  no  case  express 
an  answer  more  exactly  than  the  circumstances  warrant ;  thus,  no 
ordinary  thermometer,  even  of  those  used  in  physical  laboratories, 
will  read  correctly  nearer  than  tenths  of  a  degree.  Hence,  in  such  an 
expression  as  ;3.5''.74,  the  last  figure  does  not  really  mean  anything 
and  should  be  omitted.* 

*  It  is  suggested  in  the  text  that  to  plot  the  graph  oiy  =  kx,  the  two 
determhiing  points  may  be  chosen  as  (0,  0)  and  (1,  k).  This,  as  well  as 
the  ordinary  practice  of  plotting  ax+hy  +  c=  0,  by  locating  the  points 
.(0,  —c/b),  (  —  c./a,  0)  is  unwise  as  a  ypneral  rule.  The  disadvantages 
are  two  in  the  latter  case,  and  at  least  one  applies  to  the  former  also  : 

1.  The  numbers  that  fix  the  points  thus  located  may  be  fractions 
with   large  denominators  which  cannot  be   located  accurately ;   proper 


80]  PROPORTION  U.', 


EXERCISES   X:   CHAPTER   V 

By  the  principles  of   §  78,   solve   the   following   equations 
(Exs.  1-6): 

1.  -i^  =  -  •    [Use  II  or  V  or  VI  (2).l     4.    -^  =  -  • 
p-S     5      ■-  ^  ^-'  x+2     3 

2.  ^i±ll  =  ll.    [Use  II  or  V  or  VI  (2).]     5.    ^  =  ^. 
X  —  G      7  X     ,324 

3     ^-3^2  g     a;-2_     2 

xF         3'  ■       16   ~a;  +  2' 

Solve  the  following  equations  (Exs.  7-10)  for   the   letters 
indicated : 


-v if      j^ 

7. —  -  •      [Solve  for  x  ;  also  soh'e  for  «.] 


8.     =  -  •      [Solve  for  iv ;  also  solve  for  r.] 


9.     = : —      [Solve  for  x ;  also  solve  for  «."] 

x  —  ac-\-d 

^n     ci  —  b  —  c  c  m  ^      t         T 

10.    , =  — — -•      [Solve  for  o.] 

a  +  b  +  c      26  +  3c  -" 

If  -  =  -^  show  that  the  statements  of  Ex.  11-29  are  correct: 
b      d 
in  doing  so,  it  is  justifiable  to  clear  of  fractions  if  no  easier 
process  is  evident : 

a-  _c-  a-  —  h^  _c-  —  d-  ah      h' 

h-      d^  a-  (r  cd     d^ 


integral  points  might  be  as  easily  found,  and  much  more  easily  plotted. 
E.g.,  to  plot  accurately  the  graph  of  7  x  +  Sy  =  10  by  locating  (12,  0) 
and  (0,  3^)  is  a  practical  impossibility  ;  to  do  .so  by  locating  (1,  1)  and 
(4,  —  6)  is  exceedingly  easy. 

2.  The  points  located  as  above  are  often  too  close  together  for  accurate 
plotting.  A  slight  deviation  in  either  point  may  throw  the  general  direc- 
tion of  the  line  entirely  off.  The  graph  oi  y  =  yV^  cannot  be  accurately 
plotted  with  equal  small  scales  for  x  and  ?/,  by  locating  (0,  0)  and  (1,  y'j). 
The  work  can  be  more  accurately  done  by  locating  (0,  0)  and  (10,  1)  ;  or 
better,  (10,  1)  and  (-  10,  —  1). 

hkdkick's  kl.  ai.o.  —  10 


146  FRACTIONS  [Ch.  V 

14.    a^±b^^r±d^^  ^^ 


15.    u^r^^^iji^.  20. 


ab 

cd 

a-}-nb  _c 

+  nd 

b 

d 

pa  -\-  qb 

j)c  +  qd. 

a 

G 

a2  +  62 

ac  +  bd 

^g      i-  T  ^^^^^^    ,    ^^-^  21. 


17.     ^^^^CJi_  =  "'^^"".  22. 

ac  +  6d      c^  +  d^ 


18     «^-2ac  +  c-_qc  22 


a'-c' 

b'^-d- 

a-  +  c' 

b'  +  d' 

a  +  7ib 
a 

c  +  nd 
c 

pa  4-  qb 
ra  +  sb 

pc  -f  qd 
re  +  i'c/ 

ax  -\-  by 

a      b 

ex  +  dx 

c     d 

ab  —  c^_ 

a^  —  cd 

b^-2bd  +  d-     bd  b' -  cd     ab-d^ 

If  a,  b,  c,  d  are  all  different  from  zero,  show  that : 

24.  If  P^L+a^  =  Pl±9i,  then  ^^  =  ^ . 

a  c  b      d 

25.  If   ^'  +  ^'="^  +  ^^,then^  =  ^. 

ac  +  bd      cr  +  d^  b      d 

26.  Express  the  simple  interest,  i,  at  6  %  on  $  500  for  t 
years.  Draw  the  graph.  In  how  many  years  will  the  intei-- 
estbe$90?  $60?     What  quantities  are  here  proportional  ? 

27.  Express  the  amount  in  the  above  problem.  Draw  the 
graph  on  the  same  diagram  as  that  for  example  26.  What 
quantities  are  proportional  ?  State  and  explain  the  relative 
position  of  the  two  figures. 

28.  Compare  by  an  equation  the  readings  of  a  Centigrade 
and  a  Reaumur  thermometer  as  given  in  the  table  at  the  back 
of  the  book.  What  quantities  are  proportional  ?  Draw  the 
graph. 

Compare  the  readings  of  a  Fahrenheit  and  a  Reaumur  ther- 
mometer. What  quantities  are  proportional  ?  Draw  the 
graph. 


80]  PROPORTION  147 

29.  John  can  run  20  feet  a  second  ;  Henry,  15.  Compare 
the  distances,  x,  y,  from  the  same  starting  point  of  John  and 
Henry  at  any  time.    What  quantities  are  proportional  ?    Draw 

the  graph. 

30.  If  John  gives  Henry  a  start  of  5  feet,  compare  the 
distances  from  John's  starting  point,  in  Ex.  29.  What  quanti- 
ties are  proportional  ?  Draw  the  graph.  Explain  the  relation 
to  the  figure  for  Ex.  29. 

31.  If  John  and  Henry,  in  Ex.  30,  meet,  show  that  y  =  x  for 
the  point  of  meeting.  Draw  the  graph  giving  all  values  for 
which  this  relation  holds.  Where  will  John  meet  Henry  in 
Ex.  30  ?      Solve  this  problem  also  by  means  of  the  equations. 

32.  Try  to  find  graphically,  as  in  Ex.  31,  the  distance  at 
which  John  and  Henry  meet  in  Ex.  29.     Explain  your  failure. 

33.  State  the  general  formula  for  the  interest,  /,  in  terms  of 
the  principal,  p,  the  rate  per  cent,  r,  and  the  time,  t,  for  simple 
interest.  If  the  i)rincipal  and  time  are  constant,  to  what  is 
the  interest  proportional  ?  To  what  other  quantities  can  we 
make  the  interest  proportional  by  keeping  certain  quantities 
constant  ? 

34.  What  is  the  number  of  minute  spaces,  m,  traveled  by 
the  minute  hand  of  a  clock  in  60  minutes  ?  the  number,  h, 
traveled  by  the  hour  hand  ?  Wliat  is  the  relation  between  m 
and  h  ?  Draw  the  graph.  At  12  o'clock  what  are  the  posi- 
tions of  the  two  hands  ?  Thus,  m  and  It  may  represent  the 
positions  of  the  two  hands  m  minutes  i:)ast  twelve. 

35.  What  are  the  positions  of  the  two  hands  at  3  o'clock  ? 
If  m  and  h  are  to  represent  the  number  of  minute  spaces  dis- 
tance from  the  figure  "XII"  of  the  two  hands  at  m  minutes 
past  three,  what  equation  connects  m  and  h  ?  Draw  the 
graph. 

Find  graphically  when  m  =  1i ;  that  is,  when  the  hour  and 
minute  hands  will  be  together.     Solve  also  from  the  equation. 


148  FRACTIONS 

36.  Find  graphically  and  from  the  equation,  when,  after  6 
o'clock,  the  hour  and  minute  hands  of  the  clock  will  be  to- 
gether. 

37.  Find  graphically  and  from  the  equation,  when,  after 
three  o'clock,  the  hour  and  minute  hands  will  be  exactly  oppo- 
site one  another. 

Find  also  when  before  three  o'clock  this  will  happen. 

38.  What  is  the  total  surface  area,  A,  of  a  closed  cylinder 
whose  height  is  h  and  the  radius  of  whose  base  is  r  ?  Factor 
the  result.  If  r  is  constant,  to  what  variable  quantity  is  A 
proportional  ?  Assume  some  convenient  value  for  r,  say  r  =  7 
or  i  (take  it  =  3|),  and  plot  the  relation  between  A  and  h. 

Plot  on  the  same  diagram  the  surface  area,  B,  of  a  corre- 
sponding cylinder,  open  at  the  top. 

Plot  on  the  same  diagram  the  surface  area,  C,  of  a  cylinder 
open  at  both  ends. 

Compare  the  three  figures. 

39.  Each  of  two  boxes  has  a  square  bottom  and  rectangular 
sides.  The  heights  are  the  same.  The  first  is  2  feet  wide 
and  has  a  closed  bottom  and  top.  The  second  has  a  bottom, 
but  no  top;  and  it  is  3  feet  wide.  Find  the  common  height 
of  the  boxes  if  the  amounts  of  lumber  used  in  making  them 
are  in  the  ratio  of  8  to  13. 


PART  IV.  FRACTIONAL  EQUATIONS 

81.    Fractional  Equations.     A  fractional  equation  is  one 

which  contains  fractional  expressions. 

Equations  with  simple  fractional  coefficients  are  not  usually  in- 
cluded, because  they  can  be  so  easily  solved;  but  they  are  none  the 
less  fractional  equations.     Some  such  equations  were  solved  on  p.  .58. 

Ex.  1.   Thus,! .«  +  5  =  7  is  an  easy  fractional  equation. 

Subtract  .5  from  each  side  :  |  a;  =  2. 

Divide  each  side  by  2  :  i  x  =  1. 

Multiply  each  side  by  3  :  x  =  .3. 

2  X 1  6  a; 

Ex.  2.    A  more  typical  example  is  —  = — r  ■ 

•'^  x  +  2       Scc  +  lO 

Multiply  each  side  by  x  +  2  and  then  by  3  x  +  l»l : 
(:3  X  +  10)  (2  X  -  1)  =  6  X  (x  +  2), 
or,  6  x-^  +  17  X  -  10  =  6  x^  +  12  x. 

Subtract  6  x^  from  each  side  : 

17x-  10  =  12  X. 

Subtract  12  x  from  each  side,  and  add  10  to  each  side  : 

.5  X  =  +  10, 
or,  X  =  2. 

This  we  verify  by  trying  x  =  2  in  the  given  equation.     Thus, 

2-2-1  6-2  3     12 

- — — ,  or,  -  =  — , 

2  +  2        3-2  +  10         4     16 

which  is  true ;  hence  we  conclude  that  x  =  2  is  a  correct  answer. 

As  in  this  example,  it  is  usually  best  to  clear  of  fractions 
(§  78)  immediately.  To  do  so,  rmdlvply  both  sides  by  the 
L.  C.  M.  of  the  denominators,  if  it  can  be  found,  or  at  least 
by  as  low  a  multiple  as  is  practicable.  Or  we  may  sitnply 
cross-multiply,  as  in  §  78,  No.  V,  p.  138. 

149 


150  FRACTIONS  [Ch.  v' 

Ex.3.    -^-_+^z:i  =  2^ZL6. 
5  X  —  5      X  —  1         2  X 

The  L.  C.  M.  of  the  denominators  is  10  x  (x  —  1)  ;  multiplying  both 
sides  by  this  gives,     2  x  .  3  +  10  x  •  (x  —  4)  =  5  (x  —  1)(2  x  —  6). 

(This  is  also  found  by  cross-multiplying,  §  70,  No.  V.) 

Or,  6  X  +  10  x2  -  40  X  =  10  x^  -  40  x  +  30. 

Subtract  10  x'^  —  40  x  from  eacii  side  : 

6  X  =  30,  or,  X  =  5. 

^,     ,  3         ,5-4     2-5-6^,.    3,1       4 

Check : = ,  or,  —  +  -  =  -— . 

5-5-55-1  2-5  20      4      10 

In  case  the  answer   found   makes   any  denominator  of 
the  original  equation  zero^  that  answer  must  be  discarded,! 
since  division  by  zero  is  impossible.     (See  §  46,  p.  75  ; 
§  65,  p.  106;  §  67,  p.  109.)  I 

EXERCISES   XI  :  CHAPTER   V 

Solve  the  following  equations  for  the  vuiknown  quantities  ; 

10  A; +  35      5  A; +  5 


y-3 

y 

2y-5 

22/  +  2 

3^-4 

t  +  2 

6«-ll 

2t  +  l 

X-1 

2x^\ 

x  +  3 
2  x-  + 12 

2t-\-a_ 

2«  +  6a 

8. 


2.     —  =  - -•  9. 


10. 


IJ, 


2  A;  +  8         k-1 
a;2_^2.T-3  _x- 


3x-  +  4a;  +  l      x-\-l 

2  n  +  5  ^  6  7t  +  20 
n-2        3  «  +  4  ' 

n-2      n  +  16 


^-a  t  +  a     ,-Find/.-|  •    M-4       w+2 

32-5      6z-25  ^^     x  +  3_x±& 


'    2  2-15     4  2-33  a; +  5      a; +  9 

m- -f  4  _  m-f-1.  ^  lOo;  —  5_4  —  5a;^ 

■       2m    ~~2  ■     4a;-l  ~~l-2a;' 

7     a''  +  a  +  3^a  +  l  ^^     k  + 18  ^  A-  +  3 ^ 

'      2a +  1           2     '  ■    ^■  +  14      A--1-2* 


81-8-2]  FKACTIONAL    EQUATIONS  151 


15. 


3  c  -  20  ^  .3  c  -  12  ^Q     ^±^^_^=i 

c  —  2  c  +  3    '  '    x  +  'S     x  +  5 


16.    ^^1+1+^1^15^2.  19.    ^-P+^  +  '2P^  =  5. 

X  —  ox  p  —  6        2^  — 10 

17     _2^+?i±6=3  20     ^l±^_2i±Ai  =  2 

■    .T+1      a-  +  3       ■  '2  +  5        2  2-1-1 

82.  Other  Cases.  If  the  terms  containing  x^  and  other 
higher  powers  do  not  cancel  as  they  do  above,  the  resulting 
equations  often  may  be  solved  nevertheless.  (See  §  66, 
p.  106.) 

X.  1 .     = 

x  +  2       3  a; +  10 
Clear  of  fractious: 

62-2+  11  X  -  30  =  5x2+  10  a:. 

Subtract  5  x-  +  10  x  froui  each  side  : 

x2  +  X  -  30  =  0, 
or,  (x+ 6)(x -5)=0. 

Hence,  either  x  +  6  =  0  or  x  —  5  =  0.     See  §  66. 

Whence,  x  =  —  6  or  x  =  5. 

Check  for  x  =  5  : 

2-5-3          5-5  7     25.,,>, 
= ,  or,  -  =  —  (check). 

5  +  2        3.5  +  10         7     25  '^  ^ 

Check  for  x  =  —  6 : 

2^X-6)_-3^_Kz:^)_         -15^:^  ^^j^^^^^^ 
-6  +  2  3(-6)  +  10         -4        -8     ^ 

Thus  there  are  two  answers,  either  of  which  is  correct. 

EXERCISES    XII:    CHAPTER    V 

Solve  for  the  niiknowu  quantity  : 
^     2  a;  +  3 ^     5  a;  3     22  +  5^z  +  5_ 

cc  +  2~2.^•  +  l'  ■      z  +  l       2  +  7 

2     ^  +  3^3<  +  5  ^      n-7  ^2n-18 

•    f_3      2f-5'  '    2?i-5     3n-20" 


1.^)2  FRACTIONS  [Ch.  V 

2rt-4_3a-6  „     1-x  ,  4:X-2      ^ 

o.     — —  — — —  •  o.     — -I =  J.. 

3x  +  l_2a;  +  3 


a- 3 

a-2 

^  +  1 

2b-l 

Sb-2 

36  +  2 

X  +  3 

_3x  +  2 

9. 


5  a;         3x4-4 


7.     ^Lir^  =  ':l,i:^±'.  10.     ^  +  1  +  ^-^  =  1. 

2x-5       x+2  k+3     k-1 

11    £+1  +  ^P-1  =  8 

p- -  5 p  +  3  ^2 ir -10 p  +  15 
p-d  2p 


12. 


83.  Linear  Equations  ;  Other  Equations.  The  equations 
in  §§  81  and  82  are  distinguished  by  liaving  onli/  the  first 
power  of  X  in  their  reduced  form  after  clearing  of  fractions 
and  simplifyiiig .  Such  an  equation  is  called  a  simple  equa- 
tion, a  linear  equation,  or  an  equation  of  the  first  degree. 
(See  pp.  25,  58.) 

If  aP'  is  the  highest  power  of  x  in  the  reduced  form,  the 
equation  is  called  a  quadratic  equation  oi-  an  equation  of  the 
second  degree. 

If  a^  is  the  highest  power  of  x  in  tlie  reduced  form,  the 
equation  is  called  a  cubic  equation,  or  an  equation  of  the  third 
degree. 

\i  x^  is  the  highest  power  of  x  in  the  reduced  form,  the 
equation  is  called  an  equation  of  the  fourth  degree,  and  so 
on. 

In  general,  if  the  reduced  form  contains  x'\  but  no 
liigher  power  of  a:,  the  equation  is  called  an  equation  of  the 
nth  degree. 

In  tliese  statements  it  is  understood  that  nothing  but 
simple  powers  of  x  multiplied  by  constant  coefficients 
remain  as  terms  in  the  reduced  form. 


82-85]  FUACriOXAL   i:quatk)Xs  153 

84.  Operations.  In  this  chapter  we  shall  deal  pruici- 
pally  with  equations  of  the  first  degree  or  linear  equations. 

The  operations  of  importance  are  ; 

1.  Clearing  of  fractions  :  explained  in  §  81,  p.  149.  (See 
also  §  78,  V.) 

2.  Multiplying,  dividing,  except  by  zero,  adding,  or  sub- 
tracting by  the  same  number  on  each  side :  explained  in 
^§  o5,  65,  pp.  r)V>,  10(). 

3.  Transposing  a  term.  To  transpose  a  term  is  to  carry 
it  from  one  side  of  the  equation  to  the  other  and  change  its 
sign;  this  amounts  to  subtracting  the  term  from  each 
side.     (Compare  §  38,  pp.  58-59.) 

Thus,  if  2  X  +  4  =  8,  subtracting  4  from  each  side,  we  have  2  x  =  8 
—  4,  which  amounts  to  carrying  tlie  term  4  to  the  other  side  and 
changing  its  sign.     Similar  operations  occur  often  above. 

4.  Cancellation  of  terms.  To  cancel  a  term  that  occurs 
on  both  sides,  remove  it  {or  blot  it  out}  on  each  side;  this 
amounts  to  subtracting  that  term  from  each  side. 

The  student  may  hereafter  use  all  of  these  operations. 
Care  should  be  taken  never  to  transpose  a  factor ;  it  is  only 
terms  that  may  be  transposed. 

The  student  should  beware  of  canceling  factors  in  equor 
tions,  although  this  is  sometimes  a  perfectly  justifiable 
l)rocess.  The  danger  is  that  one  may  cancel  a  factor  that 
is  equal  to  zero,  and  that  would  be  wrong  (§§  46,  65,  pp. 
75,  106).  For  example,  although  0  •  5  =  0  •  7,  it  does  not 
follow  that  5=7.  Instead  of  trying  to  cancel  a  factor, 
carefully  use  the  principle  that  both  sides  may  be  divided 
by  any  number  except  zero.     (See  IV,  §  65,  p.  106.) 

85.  Practical  Examples.  A  few  practical  examples 
follow,  in  which  the  preceding  principles  are  applied. 


154 


FRACTIONS 


[Ch.  V 


Ex.  1.  A  cistern  may  be  filled  by  one  pipe  in  4  hours ;  by 
another  in  5  hours.  It  may  be  emptied  by  a  windmill  pump 
in  6  hours.  If  the  pump  and  both  pipes  are  running,  how 
long  will  it  take  to  fill  the  cistern  ? 

Let  X  be  the  number  of  hours  required  to  fill  the  cistern. 
1 


Then 


=  part  filled  in  1  hour. 


The   first  pipe  fills  \  the  cistern;  the  second  ^  of  it;  the   pump 
empties  \  of  it,  in  an  hour. 
Hence, 


1=1  +  1 
a;      4      5 


1 
6' 


1      17 
X      60 
Clear  of  fractions :  60  =  17  x. 

Divide  by  17  :  x  =  f  ^  =  3^^- 

Check:  in  {^  hours  the  first  pipe  fills  f ^  •  i  =  if  of  the  cistern ;  the 
second  fills  f ^  .  ^  =  jf  of  the  cistern;  the  pump  empties  f f  •  |  =  |^ 
of  the  cistern ;  hence,  in  f f  hours  the  cistern  has  in  it  \j  +  {}  —  \j 
=  i^,  i.e.  it  is  just  full. 

Ex.  2.  Two  teams  playing  baseball  in  a  league  have  the 
following  record : 


Won- 

Lost 

A 

59 

22 

B 

56 

24 

These  teams  play  a  final  series  of  ten  games  together.  How 
many  must  A  win  in  order  that  the  ratio  of  games  won  to  games 
lost  be  greater  for  A  than  for  B  ? 

Let  X  be  a  number  A  must  win  to  come  exactly  even  with  B.  Then 
A  would  lose  (10  —  .r)  of  the  final  games,  and  the  final  totals  would 
be:  A  won  59  +  x,  lost  22 +(10 -a;);  B  won  56+(10-x),  lost 
24  +  X.     If  the  ratio  of  games  won  to  games  lost  is  the  same  for  both, 

■59  +  x      _  56  +  (10  -  x) 


or, 


22  +  10  -  X 
59  +  ,r 


24 +  x 


66 


32  -  X      24  +  X 


«o]  FRACTIOxNAL    EQUATIONS  155 

Take  each  of  the  proportions  by  "  compositiou  "  (see  I,  §  78,  p.  137)  : 

91      ^     90 
32  -  X     24  +  a:* 

Clear  of  fractions  :        91  (24  +  x)  =  90  (32  -  x), 
or,  91 .  24  +  91  x  =  90  •  32  -  90  x. 

Transpose  91  •  24  and  also  -  90  x : 

181  X  =  696, 
or,  X  =  31  If. 

If  it  were  possible  for  A  to  win  3^||  games,  the  two  teams  would 
finish  equal ;  as  it  is,  if  A  wins  3  or  less,  B  will  win  ;  if  A  wins  4  or 
more,  A  will  win.  In  baseball  leagues  the  practice  is  somewhat  dif- 
ferent from  this  simple  case. 

This  example  illustrates  the  fact  that  it  is  sometimes 
not  necessary  to  know  the  answer  to  an  example  with 
any  great  exactness,  for  here  it  is  impossible  for  A  to  win 
ix  fractional  number  of  games. 

The  example  also  illustrates  the  occasional  usefulness  of 
composition  of  a  proportion,  or  some  other  of  the  processes 
on  p.  138. 

EXERCISES  XIII:  CHAPTER  V 

1.  A  cistern  is  furnished  with  two  pipes,  one  of  which  can 
empty  it  in  G  hour.s,  the  other  in  3  hours.  How  long  will  it 
take  both,  opened  simultaneously,  to  empty  the  cistern  ? 

2.  A  can  do  a  certain  piece  of  work  in  3  days ;  B  can  do  it 
in  4  days ;  A,  B,  and  C  together  can  do  it  in  li  days.  How 
long  would  it  take  C  alone  to  do  the  work  ? 

3.  A  is  15  years  old ;  B  is  25  years  old.  When  will  A  be 
I  as  old  as  B  ? 

4.  A  is  24  years  old;  B  is  52  years  old.  When  will  A's 
age  be  half  of  B's  ? 

5.  A  steamboat  is  making  G  miles  an  hour  against  the 
wind  on  a  journey  of  30  miles.  At  a  distance  of  10  miles 
after  starting  the  wind  ceases.  The  whole  trip  occupies  3 
hours  and  40  minutes.  How  many  miles  per  hour  does  the 
wind  retard  the  boat  ? 


156  FRACTIONS  [Cii.  V 

6.  A  steamboat  travels  with  the  wind  at  the  rate  of  15 
miles  an  hour,  and  returns  against  the  wind.  The  whole  trip 
is  60  miles,  and  occupies  8  hours.     How  strong  is  the  wind  ';' 

7.  What  fraction,  equal  to  ^,  becomes  equal  to  |  by  the 
addition  of  5  to  the  numerator  and  denominator  ? 

8.  What  number  can  be  added  to  both  terms  of  the  frac- 
tion I,  to  make  a  new  fraction  equal  to  f  ? 

9.  If  the  value  of  a  fraction  is  unchanged  by  adding  7  to 
both  terms,  show  that  the  value  of  the  fraction  is  unity.     (Let 

-  be  the  fraction.) 

y 

10.  If  the  value  of  a  fraction  when  2  is  added  to  both  terms 
is  the  same  as  the  value  when  4  is  added  to  both  terms,  show 
that  the  fraction  is  equal  to  unity. 

11.  What  number  can  be  added  to  2,  4,  6,  and  10  so  that 
the  resulting  numbers  will  form  a  proportion  ? 

12.  What  must  be  the  dimensions  of  a  rectangle  containing 
1000  square  centimeters,  in  order  that  the  perimeter  may  be 
266  centimeters  ? 

13.  Find  the  numbers  proportional  to  1,  2,  3,  4  that  may 
be  added  regularly  to  70,  90,  45,  60,  so  as  to  form  a  proportion. 

14.  A  certain  fraction  is  equal  to  f.  If  from  the  fraction 
obtained  by  adding  10  to  both  terms  the  fi-action  obtained  by 
adding  5  to  both  terms  is  subtracted,  the  difference  is  ^,  what 
is  the  fraction  ?     Comment  on  the  two  results. 

15.  The  secretary  of  a  club  announces  that  the  admission 
of  ten  new  members  would  decrease  by  $1  the  assessment 
required  to  pay  a  debt  of  $  20.  How  many  members  has  the 
organization,  and  what  is  the  assessment  ? 

16.  A  merchant  sells  to  three  successive  customers  one  third, 
one  fourth,  and  one  fifth,  of  a  bolt  of  cloth  ;  lie  has  left  2  yards 
more  than  one  sixth  the  original  length.  P^ind  the  original 
length  and  the  amount  sold  to  each  customer. 


85] 


SUMMARY  151 


SUMMARY  OF  CHAPTER  V:  FRACmoNS ;  COMMON  FAC- 
TORS; REDUCTION;  OPERATIONS;  PROPORTION;  FRAC- 
TIONAL  EQUATIONS;   pp.   118-156 

Pakt  I.     Common  Factoks;  Rkduction  of  Fractions,  pp.  118- 
123. 

Removal  of  Common  Factor  in  Numerator  and  Denominator  . ■ 

—  '-±-  ^  i-;  division  of  botli  terms  by  same  mimher;  multiplication 

CD      JJ 
of  both  tenus  by  same  number;  addition  (or  subtraction)  not  allow- 
able. "  §68,  pp.  118-119. 

Reduction  to  Loirest  Terms  :  remove  all  common  factors.  Exer- 
cises I.  §  m,  pp.  119-120. 

Highest  Common  Factors :  product  of  all  common  factors.  Exer- 
cises II.  §  70,  pp.  120-121. 

Practical  work  in  Factors:  i-emoval  of  all  common  factors  readily 
found.     Exercises  III.  §  71,  pp.  122-123. 

Pakt  II.     Riles  of  Op^:HATIO^',  pp.  124-136. 

Of  three  Sif/ns  in  a  Fraction  (numerator,  denominator,  irhole  fraction), 
change  any  two.     Exercises  IV.  §  72,  pp.  121-12.5. 

Addition  and  Subtraction  of  Fractions:  reduction  to  common 
denominator,  and  addition  of  new  numerators;  simple  prob- 
lems,—  common  denominator  by  inspection.     Exercises  V. 

§  73,  pp.  125-127. 

Least  Common  Multiple  :  omit  duplicate  factors. 

Choice  of  Common  Denominator :  L.  C  M.,  or  last  multiple  practi- 
cable; fornuil  rule  for  difficult  problems.     Exercises  VI. 

§  7i,  pp.  127-131. 

Product  of  Fractions :  (product  of  numerators)  -^  (product  of 
denominators).     Exercises  VII.  §  7.5,  pp.  131-133. 

Quotient  of  two  Fractions  :  multiply  by  the  divisor  inverted. 

Reciprocal  of  a  Number:  1  divided  by  that  number;  of  a  fraction, 
the  fraction  inverted;  use  in  division. 

Complex  Fractions  :  equivalence  to  division  ;  caution  to  mark  main 
horizontal.     Exercises  VIII.  §  76,  pp.  13-1-136. 


158  FRACTIONS 

Part  HI.     Proportion,  pp.  137-148. 

Proportion  :  an  equality  of  two  fractions.  §  77,  p.  137, 

If  -  =  i" ,  then 
b      d 

I.    — ^—  —  — ■ —  ("  Composition    )  ; 
h  d 

II.    'J—A  =  ^ZL^  ("  Division  ")  ; 
b  d       ^ 

III.  ^±_^  =  £_±_^  ("  Composition  and  Division")  ; 
a  —  b      c  —  <l 

IV.  -  =  -  ("  Inversion    )  ; 
a      c 

V.    ad  —  he  ("  Clearing  of  Fractions  ")  ; 

VI.    From  ad  =  be,  -  =  -,  also  -  =  - ,  etc.  Exercises  IX. 

''      ^            ^      '^  §  78,  pp.  137-138. 

Variables  in  Proportion  :  y  =  kx.  §  79,  pp.  139-140. 
Equation  y  =  kx  :  straight-line  figure. 

Example^!  in  Proportion.     Exercises  X.  §  80,  pp.  140-148. 

Part  IV.  Fractional  Equations,  pp.  149-1.56. 

Clearing  of  Fractions:  multiply  both  sides  by  the  L. CM.  Exer- 
cises XI  and  XI F.  "  §§  81-82,  pp.  149-152. 

Degree  of  an  Equation :  degree  of  the  highest  power  of  r  after  sim- 
plification ;  linear  or  simple,  first  degree  ;  quadratic,  second 
degree ;  etc.  §  83,  p.  152. 

Permissible  Operations  : 

(1)  Clearing  of  fractions; 

(2)  Multiplying,  dividing,  etc.  (both  sides)  ; 

(3)  Transposition  of  terms ; 

(4)  Cancellation  of  terms  in  equations.  §  84,  p.  153. 
Examples  stated  in  English.     Exercises  XIII.         §  85,  pp.  153-156. 


CHAPTER    VI.     SIMULTANEOUS   LINEAR 
EQUATIONS 

86.  Introduction.  Problems  in  which  two  distinct  state- 
ments are  made  lead  to  two  equations  which  should  both 
be  true.  Such  pairs  of  equations  are  called  simultaneous 
equations. 

If  the  equations  which  are  formed  are  of  the  first  degree 
(see  pp.  68,  152),  the  two  taken  together  are  said  to  form 
a  pair  of  simultaneous  linear  equations. 

Ex.  1.  Suppose  the  sum  of  two  numbers  is  45  and  their  dif- 
ference is  18.     "NMiat  are  the  numbers  ? 

Notice  that  such  an  example  may  easily  arise  in  a  practical 
problem. 

Statement  :     Let  x  and  y  he  the  two  numbers;  then 

f  X  +  ?/  =  45,  i.e.  their  sum  is  45.  (1) 

[a;  —  //  =  18,  i.e.  their  difference  is  18.  (2) 

1.  The  Figure.  The  first  equation  a7o»ehas  many  different 
.solutions:  10  and  35,  11  and  34,  12  and  33,  lOi  and  34i  etc. 
For  this  reason  the  first  equation  alone  is  called  an  indetermi- 
nate equation.  (See  also  Chapter  IX,  p.  237.)  The  correspond- 
ing graph  is  a  straight  line  since  the  equation  is  of  the  first 
degree  (§  80),  AB  in  the  figure. 

The  second  equation  is  likewise  an  indeterminate  equation 
of  the  first  degree;  its  graph  is  the  straight  line  CD  in  the 
figure. 

159 


160 


SIMULTANEOUS   LINEAR   EQUATIONS       [Ch.  VI 


Any  point  on  AB  corresponds  to  a  pair  of  numbers  whose 
sum  is  45.  Aiiy  point  on  CD  corresponds  to  a  pair  of  num- 
bers whose  difference  is  18. 

There  are  inaiiy  such  pairs;  but  there  is  evidently  only 
one  point  that  lies  on  both  lines,  i.e.  there  is  only  one  pair 
of  numbers   that  gives  the  sum  45  and  the   difference   18. 

Y 


Fig.  23. 

A  glance  at  the  figure  shows  that  a:  =31.^,  ?/=13-|  are 
about  correct ;  a  trial  of  these  numbers  shows  tliat  they 
are  exactly  correct :   31 1  +  131  =  45,  311  _  131  =  18. 

2.  Solution.  These  answers  can  be  found  otherwise 
by  several  different  methods,  one  of  which  follows ;  other 
methods  are  given  in  §  90. 

Method  I.     By  Addition  ok  Subtraction. 

■x  +  y  =  \o,  (1) 

18.  (2) 


Ex.  1. 


\x-y  = 


Srj] 


SlMrL'I'AXKOrs    LIXKAi:    KglAl'IOXS 


Itll 


We  note  that  tlie  sidiis  of  these  equal  numbers  are  equal ;  hence,  add- 
ing the  right  side.s  together  and  the  left  sides  together,  we  get  an 
equation  that  contains  only  the  letter  x : 

2  X  =  63, 
whence,  x  =  31  >,  which  agrees  with  the  figure. 

Likewise,  subtracting  the  right  sides  and  also  the  left  sides  of  the 

original  equations,  we  get  : 

2  y  =  27, 
or,  y  =  13^,  which  agrees  with  the  figure. 

Hence,  x  =  31|,  //  =  13|,  are  the  answers. 

Check :        X  +  1/  =  31^  +  lU  =  45  ;  x  -y  =  3H  -  13^  =  18. 
The  graph  should  always  I)e  drawn,  as  on  p.  160,  as  a  check  on  the 
correctness  of  the  work  done,  at  least  until  tlie  student  is  so  sure  of 
his  ability  that  there  is  small  chance  of  error. 

|.3x-  +  2/y  =  12,  (1) 

to  find  X  and  y. 


Ex.  2.    Given 


:  +  5  2/  =  20, 


(2) 


1.    Thp:  Figure. 

the  figure  : 


in  (1) 


I  if  a:  =  0,  >/ 
1  if  ^  =  U,  X 


To  draw 
=  6,     (A) 




"^ 

^^     4^ 

S     ^ 

\ 

^^  ^eT 

^>.  S4- 

^    V, 

^^-sS^A 

^T 

^^2:    M 

(3        .?\x 

X           \-X 

S     ^v 

\    -t 

^ 

s: 

1 

1    1    1 

Fig.  24. 


^       [  a  tj  =  0,  X  =  5.    (C) 

Drawing  these  points,  we 
find  the  lines  AB  and  CD  to 
represent  equations  (1)  and 
(2),  respectively.  The  only 
common  point  P  gives  x  =  1.7, 
y  =  2.8,  about.  Let  us  try  to 
solve  and  find  the  answers  exactly. 

Method  T.     By  Addition  or  Subtraction. 

rSx+2y  =  12,  (1) 

U2-  +  5y  =  20.  (2) 

Tn  order  to  obtain  an  equation  that  contains  only  the  letter  y, 
multiply  both  sides  of  (1)  bv  4, 

(3)  12.c  +  8//  =  48, 
and  multiply  both  sifles  of  (2)  by  3, 

(4)  12  j;  +  loy  =  60. 
hedrick's   el.   alg. 11 


162 


SIMULTANEOUS   LINEAR   EQUATIONS         Ch.  VI 


Subtract  (3)  from  (4)  : 

72/ =  12, 

2/  =  V  =  If- 
In  actual  work  we  write  this  as  follows : 


3  X  +  2  </  =  12 

ix  +  5// =  20 


7^  =  12 
y  =  lf 
The  numbers  —  4  and  3  indicate  the  multipliers;  notice  that  —  4  is 
written  in  place  of  +  4,  in  order  that  we  may  add  instead  of  subtract. 
Likewise,  we  find  an  equation  that  contains  only  x : 


5 
_  2 

Sx  +  2y=12 
4a; +  5?/ =  20 

whence. 
Hence, 

Check: 

7x           =20 

X           =  2f . 
X  =  2f ,   y  =  If. 
3(2f)+2(U)-6«  +  24_84_i2^ 

4(2 

?)|5(l5)-80  +  «0_140_2o_ 

V  t 

Notice   that  the   answers   obtained  from   the  figure  are  slightly 
incorrect : 

a:  =  2f  =  2.8.555  •••  (we  found  2.8  in  the  figure). 
y  =  1^  =  1.714  •••    (we  found  1.7  in  the  figure). 
We  might  also  find  x  after  finding  y  by  putting  the  value  of  y  found 

(y  =  If)  in  place  of  x  in  either  of 
the  given  equations;  thus,  putting 
y  =  If  in  (1),  we  find  3  x  +  2  (1?) 
=  12,  whence  x  =  2f .  But  it  is 
better  to  do  the  work  as  shown, 
because  a  mistake  in  finding  y 
does  not  then  cause  an  error  in 
the  value  of  x  also. 


Ex.  3. 


4  .^•  +  2  2/  -  5  =  0. 

Transpose  the  terms  to  get  the 
ec^uation  in  the  form 
3x-  y  =  8, 
4  a:  +  2.V  =  5. 
The  figure  is  as  shown.  The  results  are  therefore  uhuut  x  =  2,  ^  =  —  1 1. 


.sii-b-J  .SIMULTANEOUS   LINEAR   EQUATIONS  163 

The  solution  (Method  I,  by  addition  and  subtraction)  is 

4  jr  +  2  //  =  -T  3  I  4  X  -f  2  y  =  5 


10.r  =21  10y  =  -\7 

X  =2.1  y  =  -l-7 

CAect;     y  =  3  X  -  8  :   -  1.7  =  3(2.1)  -  8  =  6.3  -  8  (correct). 
4x  +  2y-5=0:  4(2.1)  +  2(  -  1.7)  -5  =  0, 
or,  8.4  -  3.4  -  5  =  0  (correct). 

Notice  that  the  answers  from  the  figure  are  not  precisely  correct ; 
they  agree  with  the  precise  result  as  nearly  as  we  could  expect. 

87.  Formal  Rule.  Elimination.  The  method  of  solu- 
tion just  given  may  be  summed  up  in  the  following  rule : 

Multiply  both  sides  of  each  equation  by  the  coefficient  of  y 
[or  of  x]  in  the  other  equation ;  subtract  the  two  resulting 
equations. 

The  new  equation  has  no  term  in  x  [or  j/]  ;  solve  it  for  y 
\or  x'\.     Proceed  similarly  to  find  x  [or  y~\. 

Any  process  which,  as  above,  results  in  g-etting  rid  of 
one  of  the  unknown  quantities,  is  called  elimination.  AVe 
say  that  y  (for  example)  has  been  eliminated  when  we 
find  a  new  equation  which  is  free  from  y.  The  purpose 
of  eliminating  one  unknown  quantity  is  to  find  an  equa- 
tion in  one  letter  alone.  Such  an  equation  can  be  solved 
by  previous  rules. 

In  equations  that  contain  fractions  we  first  clear  of 
fractions,  then  proceed  as  above. 

EXERCISES   I:     CHAPTER   VI 

[In  each  of  the  following,  draw  the  graph,  and  estimate 
solutions;  then  solve  by  addition  and  subtraction.] 

1.  The  sum  of  two  numbers  is  19;  their  difference  is  6. 
What  are  the  numbers  ? 

2.  The  sum  of  two  numbers  is  —  10,  their  difference  is  2. 
What  are  the  numbers  ? 


1(34 


SIMULTANEOUS   LINEAR   EQUATIONS        [Cii.  VI 


3.  Divide  20  into  two  parts,  one  of  which  shall  be  four 
times  the  other. 

4.  Divide  35  into  two  parts,  which  shall  be  in  the  ratio  of 
3  to  2. 

5.  What  fraction  becomes  equal  to  |  if  each  term  is  de- 
creased by  1,  and  to  f  if  each  term  is  increased  by  1  ? 

6.  Twice  the  difference  of  two  numbers  is  12;  twice  the 
greater  number  exceeds  the  lesser  number  by  20.  What  are 
the  numbers  ? 

7.  Half  one  number  exceeds  one  third  another  number  by 
unity ;  the  first  number  is  less  than  the  second  by  unity. 
What  are  the  numbers  ? 

8.  What  number  of  two  digits  exceeds  7  times  the  sum  of 
its  digits  by  3,  and  exceeds  16  times  the  difference  between 
the  tens'  and  units'  digit  by  4  ? 

Solve  the  following  pairs  of  equations  for  the  letters 
appearing  in  them : 


10. 


11. 


12. 


13. 


14. 


21. 


\p  +  Sq  =  9. 

[3x-10t  =  S. 

{2  m  —  n  =  1, 
[6  n- 11m  =7. 

r4a;  +  3?/  =  12, 
jo  a; -14.?/ =  27. 

f  _?/  =  5  cc  +  3, 


15.    { 


r8.9  +  5^'  +  l  =  0, 


x  =  2y 


24. 


4fc 

6  r  - 


3r  =  l, 
2  A-  =  1. 


16. 


17. 


18. 


19. 


20. 


[  4  s  =  10  I-  +  7. 

r  /  -  5  «  =  4  /  -  fi  n, 

|z_(5„_/)  +  (/  +  4)  =  0. 
r  _  3  H  +  8  V  =  5, 
^u  +  v=2. 

(6-.x-)-(?/  +  3.r)  =  -l, 
{x-\-y)  +  {\x-^y)  =  12. 

2x-y  =  l. 


■h'' 


'll-3y  =  3. 


'i-^=l 

3      5        ' 
\s-t  =  l. 


22. 


^  +  ^  =  3, 

7      6 

— i a  =  15. 

9  ^ 


23. 


4      12' 
3^^1 

16      2' 


87-88] 


SLM  [ : I/l' A  N  KOrs    LI  X E A  It    KC^ l'  AT I(  )NS 


1«J5 


Solve  the    following   for   the    letters    indicated;    draw   the 
figure  for  any  convenient  choice  of  the  other  letters: 


24. 


25. 


26. 


28. 


29. 


x-]ry  =  a, 

x-y  =  b.     {x,  y.) 

((j>  —  Ixj  =  0, 
^2'  +  "7  =  "'  +  ^'• 
ax  +  by  =  2  xy, 
bx  —  ay  =  .r  —  y''. 

a  +  {n  —  l)(l  =  l, 

,  n(7i—  1)  , 
na  +  -^—- — ^  a  = 


(a,  b.) 


27.    VJ:-!1  =  2, 
x      y 

m       ji^  _  ^ 

Wx      2y~    ' 

[SuGGKSTiox.      Fh-st  solve  for 

—  and  -   ;is  if  for  single  letters; 
X  y 

then  solve  for  ?«  and  h.] 


s.    (a,  d.)    (Compare  pp.  325, 326,  §§  157, 
158.) 


{k  +  l)x  +  ?>y  =  m, 

mx  —  2y  =  k.     (_x,  y ;  A-,  m.) 


30.    ax  +  by  =  mr, 

bx  —  ay  =  nr.     (x,  y;  a,  b;  x,  r;  b,  r.) 

88.  Impossible  Case.  If  a  problem  contains  two  state- 
ments that  are  contradictory,  the  simultaneous  equations 
formed  cannot  be  solved.  ^ 

Ex.  1.  Find  the  dimensions  of  a  rectangle 
whose  perimeter  (the  entire  boimdary)  is  10 
and  whose  two  dimensions  have  a  sum  7. 


a; 


Fig.  L'G. 


Let  us  try  to  draw  the  figure ;  let  x  and  y  be  the 
sides.     Then  the  perimeter  is  2  x  +  2  ^,  and 

(1)  2  X  +  2  y  =  \0. 
But 

(2)  x  +  y^7, 

since  the  sum  of  the  two  sides  is  to  he  7. 

These  two  statements  are  contradictory  for,  f  i-oni  the  first  equation, 
when  both  sides  are  divided  by  2,  x  +  y  =  5. 
Now  a;  +  !/  cannot  be  both  5  and  7,  hence  there  is  no  answer. 

We  could,  however,  find  many  different  pairs  of  numbers  that 
satisfy  only  one  of  the  two  given  equations.  In  fact,  (1)  by  itself 
is  an  indeterminate  equation  of  the  first  degree ;  it  therefore  has  a 
straight-line  graph,  the  lower  line  in  Fig.  27. 


166 


SIMULTANEOUS   LINEAR  EQUATIONS        [Ch.  VI 


W, 


(5,-0)- 


^<)> 


Fig.  27. 


Likewise,  (2)  is  represented  by  the  upper  straight  line.     Now  these 

lines  never  cross ;  ior  if  they  did, 
their  common  point  would  be  on 
both,  i.e.  there  would  be  a  pair 
of  numbers  whose  sum  is  5  and 
whose  sum  is  also  7,  which  is 
absurd. 

Note  1.  The  kind  of  argu- 
ment just  used  is  often  called 
reductio  ad  absurdum,  or  reduction 
to  an  absurdity ;  we  prove  that  the 
statement  made  (in  this  case  the 
statement  that  (1)  and  (2)  do 
not  cross)  is  true  by  showing  that 

otherwise  an  absurd  (incon-ect)  conckision  would  follow. 

Note  2.     Straight  lines  in  the  same  plane  that  never  cross  are 

called  parallel  lines. 

Two  simultaneous  equa- 
tions in  two  unknown  quan- 
tities that  correspond  to  par- 
allel lines  in  the  figure  have 
no  pair  of  solutions. 

A  pair  of  numbers  that  are 
the  solutions  of  a  pair  of  simul- 
taneous equations  corresponds  to 
a  point  on  both  lines. 

5,         (1) 


Ex.  2. 


9  a; 


6.K 


32/  =  -10.    (2) 

Here  it  is  not  easy  to  see  by 
mere  inspection  that  the  equa- 
tions are  not  solvable.  But  if 
we  soloe  for  y  in  each  one,  we  find 

f2/  =  .3x-|,        (4),  from  (1) 

|y/  =  3a;  +  V,      (.^)).  from  (2) 

These  are  represented  by  parallel  lines,  for  each  one  is  parallel  to 
the  line. 
(6)  y^'i  JC. 


Ex.3. 


88]  SIMULTANEOUS   LINEAR   EQUATIONS  167 

lu  fact,  (5)  is  formed  by  raising  (0)  vertically  by  Y>  while  (4)  is 
formed  by  lowering  (0)  vertically  by  =,.  (See  p.  25.)  It  is  clear  that 
the  three  lines  are  parallel,  i.e.  tliat  they  never  meet. 

Notice  that  if  one  does  not  suspect  that  a  pair  of  equations  has  no 
pair  of  solutions,  attention  will  be  called  to  the  suspicious  character 
by  a  figure.  If  a  figure  shows  the  lines  fairly  parallel,  we  work  as 
above. 

The  line  y  =  k.r  +  c  is  parallel  to  the  line  y  —  kx;  hence 
two  lines  are  parallel  if  the  coefficient  of  x  is  the  same  in 
each  after  solving  for  y  in  each. 

An  attempt  to  solve  two  such  equations  will  call  attention  to  the 
peculiar  nature  of  the  example. 

f22/  =  6x-5,  (1) 

■  9x--3t/  =  -10.  (2) 

We  should  transpose  terms  to  get  them  in  the  form  below ;  the  solu- 
tion would  then  be 

-3|6x-2?/  =  5  (3),  from  (1) 

2  I  9  a;  -  3  y  =  -  10  (4),  the  same  as  (2) 

0  +     0  =  -  35 

This  is  absurd ;  —  35  cannot  be  equal  to  zero.  If  we  did  not  suspect 
that  there  was  no  solution  before,  such  an  absurd  result  should  make 
us  carefully  investigate,  as  above. 

Notice  that  elimination  of  either  letter  also  eliminates  the  other. 


EXERCISES   II:  CHAPTER   VI 

In  each   of   the    following   plot  the  equations ;  solve   each 
equation  for  one  of  the  unknown  letters  and  compare  results : 

^      i3y  =  Qx  +  2,  ^     \2x-y=-, 


p-29  =  7, 

r  3  »  +  4  r  =  6, 

U2  r  -  w)  -  -  («  +  3r)  + 12  =  0. 


168 


SIMULTANEOUS   LINEAR   EQUATIONS        [Ch.  VI 


5.  Find  two  integers  whose  sum  is  12,  such  that  the  sum  of 
the  integers  next  following  them  is  18. 

6.  What  fraction  becomes  equal  to  f  if  its  terms  are  either 
each  increased  by  5  or  each  decreased  by  2? 

89.  Equivalent   Equations.      A  problem  may  lead  to  two 
equations  which  are  essentially  the  same  one. 


Ex.  1. 


2y  =  6x-4:, 
9x-3y-Q  =  0. 
Solving  for  y  in  each,  we  get 

3/  =  3x-2, 


[y 


Sx-2. 


(1) 
(2) 

(3),  from  (1) 
(4),  from  (2)  ^ 

In  such  a  case  any   pair    of   numbers  that  satisfies   one 
equation  also  satisfies  the  other.     The  equations  are  called  ' 
equivalent,  for  one  may  be  reduced  to  the  other  by  opera- 
tions which  are  correct. 

The  figure  in  this  case  is  the  same  for  both  equations  ; 
it  is  a  straight  line,  since  each 
equation  is  of  the  first  degree,  i 
There  are,  of  course,  an  indefi- ' 
yiitely  large  number  of  pairs  of 
numbers  that  satisfy  both  equa- 
tions^ for  any  point  on  the  straight 
line  gives  such  a  pair. 

To  test  for  this  kind  of  problem,  we 
soh-e  for  y  in  each.  If  the  result  is 
precisely  the  same  in  each  case,  we 
know  the  equations  are  equivalent. 
Attention  will  be  called  to  such  a  case 
(just  as  in  §  88)  by  drawing  a  figure, 
or  also  by  an  attempt  to  solve  by  the 
Fig.  29.  ordinary  method. 

It  is  well  to  notice  that  if  two  straight  lines  have  two 
points  in  common,  they  are   the   same    line;  i.e.  if   two 


8!?-t>yj  .SLMULTAXEULb    LiXKAll    Et^LATlONS  169 

equations  of  the  first  degree  have  two  pairs  of  solutions, 
they  are  equivalent  equations,  and  therefore  have  an  in- 
definitely large  number  of  pairs  of  solutions. 

Two  simultaneous  equations  of  the  first  degree  have 

(one  solution       \  r intersecting  lines, 

no  solution         I  if  the  two  lines  are  J  2)arallel  lines  i . 

many  solutionsj  [the  same  line         J 

EXERCISES    III:    CHAPTER   VI 

In  eaoli  of  tlie  following,  plot  tlie   equations  ;    solve  each 
equation  for  one  of  the  unkuowns  and  compare  the  results: 

\l0i/  =  6x- 2.  2.    \  3      '> 


r<)/  +  L'  =  L>(r.-m), 
3.     {    .j^  8-/ 


[4j>-6(/  =  60. 


r  ax  —  ^//  =  a-  +  b', 

4.     i  .,.  _  a  -   y  +  h      ^      (Solve  for  x,  y.) 

I  ■  —  '^—^ —  ^  U. 

[     h  a 

X  =  10  ?/  -h  3, 

(3x-25y)-5(7/  +  2)  =  -l. 

6.  Twice  the  sum  of  two  numbers  is  12 ;  the  difference  be- 
tween the  numbers  lacks  twice  the  smaller  number  of  being 
equal  to  6.     What  are  the  numbers  ? 

7.  A  number  of  two  digits  is  equal  to  four  times  the  sum  of 
its  digits;  if  the  digits  are  reversed,  the  new  number  is  equal 

I  to  seven  times  the  sum  of  the  digits.     What  is  the  number  ? 

8.  What  fraction  becomes  equal  to  |  if  its  terms  are  each 
1  diminished  by  1 ;  or  also  if  its  denominator  is  diminished  by  7 

and  its  numerator  by  5  ? 


170  SIMULTANEOUS   LINEAR  EQUATIONS       [Ch.  VI 

90.  Solution  by  Substitution.  The  problems  given 
above  may  be  solved  by  other  methods ;  one  of  these  is 
the  method  called  solution  by  substitution.  In  this 
method,  also,  we  first  eliminate  one  of  the  unknown  let- 
ters by  a  somewhat  different  process  illustrated  in  the 
following  examples,  and  then  solve  for  the  one  which 
remains,  as  before. 

Ex.1.    (Ex.l,p.l60.){_-^^^3  (2; 

Solve    (1)    for   y. 
(3)  y  =  45  —  X. 

Substitute  the  value  found  (45  -  x)  in  the  place  of  y  in  (2)  : 
X  -  (45  -  x)  =  18, 
or,  2  X  -  45  =  18. 

Solving  for  x,  we  have  x  =  31|. 

Put  this  value  of  x  in  place  of  x  in  (3)  : 

2/  ^  45  -  31i  =  13^. 
These  are  the  answers  found  above  (p.  161). 
Check :  x  +  y  =  31i  +  13i  =  45  (correct). 

x-y  =  Sll  -  13i  =  18  (correct). 

r3x-\-2y  =  12,  (1) 

Ex.2.    (Ex.2,p.l61.){^^^.^^^^,^  ^,^ 

Solve  (1)  for  >i : 

Substitute  for  y  m  (2)  : 

4.  +  5(li^)=20. 

Clear  of  fractions  :  8  x  +  60  -  15  x  =  40. 

Simplify:  7  x  =  20. 

Solve  for  X :  a;  :=  2f . 

Substitute  2^  for  x  in  (3)  : 

12-3(2f)^-^^-y_24_12_^, 
y  2  2  14      7 

These  are  the  answers  found  above  (p.  162). 


90]  SlMULTANKOl'S    LINEAR   EQUATIONS 

60  +  2i 


171 


Check :   3  x  +  2  //  =  3 ( Y)  +  -^  (  ¥)  ■ 
4a;  +  5^  =  -l(-V°)+")(V)^ 


7 
80  +  60 


Ex.  3.   (Ex.  3,  p.  162.) 


y  =  3x-S, 


=  12   (correct). 

=  20   (correct). 

(1) 

=  0. 

(2) 

4x  +  2y-o  =  0. 

Substitute  (3  x  —  S)  for  //  in  (2)  : 
(3)  4  X  +  2  (3  X-  -  8)  -  5  =  0, 

or,  10x-16-5=0, 

or,  I0x  =  21, 

or,  X  =  2.1. 

Substitute  2.1  for  x  in  (1): 

y  =  -d  (2.1)  -  8  =  6.3  -  8  =  -  1.7. 

These  are  the  answers  found  above  (p.  163). 

Check :         y  =  Sx~8;    -  1.7  =  3  (2.1)  -  8   (correct). 

ix  +  2y-5  =  0;  4(2.1)+  2(- 1.7) -5  =  8.4  -  3.4  -  5  =  0  (correct). 

This  last  example  is  easier  by  this  metliod  than  by  the  previous 
method. 

EXERCISES   IV:    CHAPTER   VI 

Solve  the  following  by  substitution ;   in  each  case  draw  the 
figure : 


4. 


x  +  yz=17, 
•ix-5y  =  n. 

6x  +  9t  =  5A, 
5x-3t=3. 


2a-~oh  =  9, 
a  +  3h^-\. 


3. 


ly+9n  =  8, 
on  —  3y=\. 


7. 


f  aii  -\-  hm  =  0, 
5.    'I 

I  am  —  bu  =  I  (m^  +  n-). 

(Solve  for  a,  b.) 

g_    |3.T  +  4?/  =  5, 
[Sx  +  9y  =  10. 


8.     \ 


t  5       14 

f  X-      3^  _  3 
'  6      4~2' 


4     6     12 


172 


SIMULTANEOUS   LINEAR   EQUATIONS       [Ch.  VI 


91.  Solution  by  Comparison.  A  third  method  of  solu- 
tion, called  solution  hy  comparison,  is  illustrated  by  the  fol- 
lowing examples.     This  is  also  a  process  of  elimination. 


Ex.  1. 


f  a;  +  y  =  45, 


(1) 
(2) 


[  a;  -  i/  =  18. 
Solve  each  equation  for  y : 

2/  =  45  -  2-,  from  (1),  (3) 

!/  =  X  -  18,  from  (2).  (4> 

Since  45  —  x  and  x  —  18  are  each  equal  to  y,  they  are  theniselve? 
equal,  for  they  are  the  same  number  as  y ;   hence, 

45-z  =  x-  18. 

Solve  for  x:  x  —  31^. 

Substitute  31|  for  x  in  (3)  :         y  =  45  -  31i  =  13i. 

These  are  the  answers  found  above  (p.  161). 


Check . 


Ex.2. 


X  +  //  =  31^  +  13.1  _  45  (correct). 
X  -  //  =  3H  -  131  =:  18  (correct). 


^3x  +  2y  =  12, 

4:x  +  5y  =  20. 

each : 

fx-l--2.y 

20-  5m 

12  -  2  (/  _  20  -  5  .y 

3 

4 

Hence, 


Solving  this  fractional  equation  for  y,  we  find 


y 


12   —    -It 


Substituting  Y  ^oi"  V  "'  C'^)'  ^^  ^^^^ 


12  -  2(V) 


K  _  ^  _  60  _  20  _  oe 

■7* 


(1) 

(2) 


(3) 
(4) 


3  3  21       7 

Hence,  z  =2^,  y  =  If,  which  are  the  answers  found  above  (p.  lfi"2). 


91-U2J  SIMLLTANKOUS    LINKAK    EQUATIONS  173 

EXERCISES    V:     CHAPTER   VI 

Solve  the  following  by  comparison.     Draw  the  graphs: 

6  r  -  s  4- 1  =  0, 


1. 


2. 


3. 


.  3  X  +  5  //  =  25. 

2m -3  n  =  2, 
5  m  +  ?;  =  56. 

p  —  aq  =  k, 

ap  +  q^l      (For p,  q-) 


5. 


6. 


10r-3.s  +  ll  =  0. 
4  .r  +  y  =  1, 

3      5       ' 


2     10 

92.  Three  Unknowns.  Problems  in  which  three  un- 
knowns occur  can  be  solved  similarly.  We  shall  illustrate 
by  an  example  the  method  of  (1)  addition  or  subtraction^ 
which  is  most  generally  used. 

f3a;  +  ^V-2^  =  ll,  (1) 

Ex.  1.    \2x  +  3y  +  z  =  l^,  (2) 

\x  +  3y-2z  =  l.  (3) 

The  solution  by  addition  or  subtraction  is  as  follows : 


1 

.Sj:  ^ 

>/-'2z  =  11 

2 

'2x  + 

5  V  +      c=  If) 

(1) 


a:  +  3  // 


2z=  11 
■2z=     1 


(1) 


7  X  +  7  //  =49  2x  -2y 

or,        X  +     y  =7  or,     x  —     y 

Taking  the  resulting  two  equations,  we  have 

X  +  }/  =    7  1  \x  +  y 

X  —  y  =    5  —  l\  X  —  y 


=  10 
=    5 


2x=12  22^  =  2 

or,  X         =  6  y  =  i 

Substituting  6  for  x  and  1  for  y  in  (1),  we  have 

18+  1  -  2z  =  11,  orz  =  4. 

Check  :         .3  x  +  ;/  -  2  ^  =  18  +  1  -  8  =  1 1  (correct). 

2x+3^  +  2  =  12+3  +  4=19  (correct). 

X  +  dy  -2z  =  6  +3-8  =  1  (correct). 

In  this  work  we  may  group  equations  in  pairs  in  any  other  way  we 

please.     Thus, we  may  take  (1)  with  (2)  and  eliminate  y;   (2)   with 


17i  SIMULTANEOUS   LINEAR   EQUATIONS       [Ch.  VI 

(3)  and  eliminate  y  ;  the  two  resulting  equations  contain  only  x  and 
2,  for  which  we  can  solve.  In  any  case  the  purpose  is  to  reduce  this 
problem  to  one  that  we  are  already  able  to  solve,  by  eliminating  one 
letter,  thus  leaving  two  equations  in  two  letters,  which  we  solve  by 
the  preceding  methods.     The  work  when  y  is  eliminated  follows  : 

3    3a:+     y-2z^n     (1)  l|2x  +  3?/+     2  =  19     (2) 

-  1       a:  +  3  y  -  2  2  =    1     (3)      -  1  |     .r  +  3  y  -  2  z  =    1     (3) 


8  X  -  4  2  =  32  X  +32  =  18 


or, 


2x-     2=8 
X  +  3  2  =  18 


2x  -  2  =    8 
x+32  =  18 


7  X  =42  7  2  =  28 

X  =6  2=4 

Hence,  by  (1 )  3  •  6  +  y  -  2  •  4  =  11,  or  y  =  1. 

These  are  the  answers  just  found. 

Let  the  student  solve  the  same  problem  by  pairing  (1)  and  (3) 
and  (2)  and  (3),  eliminating  x  in  each  pair,  and  solving  the  resulting 
pair  for  y  and  z. 

Equations  in  more  than  three  letters  are  solved  upon  the  same 
plan ;  such  equations  will  rarely  occur  in  practical  problems  on 
topics  now  known  to  the  student,  but  they  sometimes  come  up  in 
more  advanced  work. 

EXERCISES  VI:  CHAPTER   VI 

Solve  the  following : 

r  2  a:  -  3  ?/  +  5  2  =  15,  f  3  a;  +  ?/  -  z  =  5, 

1.    ]x  +  2y-z  =  i,  4.       2cc-2/4-3  2;=13, 

[5x-y  +  ^z=l%  [x  +  2y-2z  =  0. 

f  3  a  +  6  -  7  c  =  9,  Tv  +  2  =  8, 

2        6a-2&-c  =  9,  5.    \z  +  x  =  12, 

[  3  a  +  4  &  -  10  c  =  9.  I  .X-  +  ?/  =  10. 

{x-y  +  t  =  ^,  {y  +  z-x^\, 

3.    J  2cK  +  5?/  +  3?  =  l],  6.    I  2 +  .>;-?/ =  3, 

\&x  +  lly  —  5t=9.  [x^  y  —  z  =  ^. 

[  P  -f  g  +  ?•  =  T, 

7.       /j  +  2g  +  3/-  =  10,      ^ 
[2i)  +  3ry  +  6r=:15. 


02-93] 


SLMULTAXEOUS   LINEAR   EQUATIONS 


175 


10. 


'x  +  2y  +  3z-\-t=ll, 
5  X  4-  3  ?/  -  2  2  +  2  <  =  o, 
—  X  +  >/  +  A  z  —  t  =  6, 
2  X +  4:  i/-z  +  :it  =  11. 

y  +  z  —  3  x  =  a  —  b  —  c, 
z-\-x  —  ?>y  =  h—c  —  a, 
X  -\-  y  —  3  z  =  c  —  a  —  b. 


[Solve  for  x,  y,  z\  also  for  a,  b,  c] 


93.  Practical  problems  are  given  below  which  lead  to 
simultaneous  equations.     No  new  principle  is  involved. 

EXERCISES   VII  :  CHAPTER   VI 

Solve  the  followiug  problems  by  the  use  of  two  or  more 
unknown  quantities : 

1.  A  vessel  goes  downstream  at  the  rate  of  6  miles  an  hour, 
and  upstream  at  the  rate  of  4  miles  an  hour.  What  would  be 
the  rate  of  the  vessel  in  still  water,  and  what  is  the  rate  of  the 
current  ? 

Let  s  —  rate  of  the  vessel  in  still  water,  in  miles  per  hour. 
c  —  rate  of  the  current,  in  miles  per  hour. 

Then,  s  +  c  =  6,  s  —  c^i. 

Hence,  on  solving,  s  =  5,  c  —  \. 

The  vessel  would  make  5  miles  per  hour  in  still  water,  and  the 
current  flows  at  the  rate  of  1  mile  per  hour.  These  results  evidently 
check. 

2.  A  boat  can  be  rowed  by  its  crew  12  miles  an  hour  with 
the  current;  but  only  3  miles  an  hour  against  the  current. 
Find  the  speed  of  the  current,  and  the  rate  at  which  the  boat 
can  be  rowed  in  still  water. 

3.  A  boat  is  rowed  for  3  hours  with  the  current,  and  then 
rowed  back  for  3  hours,  the  total  distance  thus  covered  being 
27  miles ;  it  is  then  allowed  to  drift  for  an  hour,  and  rowed 
with  the  current  another  hour,  thus  covering  9  miles.  What 
is  the  rate  of  the  boat  in  still  water,  and  how  swift  is  the 
current  ? 


176'  SIMULTANEOUS   LINEAR   EQUATIONS       [Ch.  VI 

4.  A  man  has  14  coins,  all  dollars  and  quarters,  amounting 
to  $  8.     How  many  of  each  denomination  has  he  ? 

5.  Ten  bills  of  denominations  $  2  and  $  5  amount  to  $  29. 
How  many  of  each  denomination  are  there  ? 

6.  I  want  the  equation  ax-\-by  =  l  to  be  satisfied  by  x  =  i, 
?/  =  l,  and  also  by  x  =  —  3,  t/  =  — 4.  How  must  a  and  b  be 
chosen  ? 

If  the  equation  is  to  be  satisfied  by        a-  =  1,  ?/  =  1, 
then,  a  +  b  =  I, 

and  if  hj  x  —  —S,  y  =  -  4,  then,  —  3 a  —  4 6  =  1. 
Solving,  we  have  a  =  5,  i  =  —  4. 

■  The  equation  is  5  x  —  4  ?/  =  1. 

Check:  5.(1)-4(1)=5-4  =  L 

5(_3)_4(-  4)=-  15  +  10=  1. 

Determine  I  and  b,  and  write  the  exact  form  of  the  equa- 
tion y  =  lx  +  b  if  it  is  to  be  satisfied  by : 

7.  x  =  2,y  =  ll;  and  x  =  —  l,y  =  2. 

8.  x  =  l,  y  =  5',  and  x=2,  y  =  3. 

9.  x=:0,  y  =  0;  and  x  =  m,  y  =  n. 

Determine  a  and  b,  and  simplify  the  equation  ax~  +  by"  =  400 
if  it  is  to  be  satisfied  by : 

10.  X  =  12,y  =  16;  and  x  =  - 16,  y  =  12. 

11.  X  =  3,  y  =  o;  and  x  =  l,  y  =  —  1- 


T> 


^  Fig.  30. 

If  two  weights,  w  and  W,  balance  when  placed  on  opposite  sides  of 
a  lever  at  distances  d  and  D,  respectively,  from  the  single  intermediate 
point  (known  as  the  fidcrum)  on  which  the  lever  rests,  then  it  is 
known  that  dw  =  DW. 


!):!]  SLMrL'I'ANKOUS    LINEAR    EQUATIONS  177 

12.  Weights  of  12  and  18  pounds  are  fastened  to  the  ends 
of  a  ten-foot  pole.  Where  must  the  pole  be  supported  in  order 
that  the  weights  may  balance  ? 

13.  An  unknown  weight  3  inches  from  the  fulcrum  of  a 
lever  is  balanced  by  another  unknown  weight  9  inches  from 
the  fulcrum  ;  an  addition  of  2  pounds  to  the  first  weight  neces- 
sitates the  removal  of  the  second  weight  8  inches  farther  from 
the  fulcrum  in  order  to  preserve  the  balance.  What  are  the 
two  weights  ? 

14.  Two  unknown  weights  balance  when  placed  6  and  9 
inches  from  the  fulcrum  of  a  lever ;  if  their  positions  are  re- 
versed, 2  pounds  must  be  added  to  the  lesser  weight  to  restore 
the  balance.     What  are  the  weights  ? 

15.  An  8-pound  weight  is  placed  at  an  unknown  distance 
from  the  fulcrum  of  a  lever ;  on  the  opposite  side  is  placed  an 
unknown  weight.  In  order  to  make  the  lever  balance  we  may 
either  increase  the  unknown  weight  by  1  pound  and  station 
it  3  inches  from  the  fulcrum,  or  else  station  it  4  inches  from 
the  fulcrum  and  add  2  pounds  to  the  other  weight.  Find  the 
unknown  weight  and  the  distance  from  the  fulcrum  to  the 
other  weight. 

16.  The  sum  of  the  three  angles  of  a  triangle  is  180°.  What 
are  the  acute  angles  of  a  right-angled  triangle  if  one  is  twice 
the  other  ? 

17.  Determine  the  three  angles  of  a  triangle  if  the  sum  of 
the  first  and  second  is  twice  the  third,  and  the  sum  of  the  first 
and  third  is  the  second. 

18.  The  equation  or +  y- =  ax-{-by  +  c  is  to  be  satisfied  by 
cc  =  4,  ^  =  2  ;  .*■  =  4,  1/  =  -  4  ;  and  x  =  5,  y  =  l.  Determine  a, 
b,  and  c. 

19.  A  boatman  in  2  hours  rows  a  certain  distance  up  a 
stream  where  the  rate  of  the  current  is  known  to  be  2  miles  an 
hour ;  he  then  rows  back  to  a  place  1  mile  beyond  the  starting 
point  in  1  hour.  Find  the  distance  he  rows  each  way,  and  the 
rate  of  the  boat  in  still  water. 

hkdrick's  el.  alg. — -12 


178 


SIMULTANEOUS   LINEAR   EQUATIONS        [Ch.  VI 


REVIEW   EXERCISES 

Solve   the  following   sets   of 
quantities : 

nx  +  5y  =  3, 
■     1.  2  a;  +  3  y  =  4. 

5a-76-c  =  16, 

2.  ■  3  a  =  2  6  +  2  c  =  10, 
,2a  +  6  +  3c  =  6. 

(af  +  y'  =  13, 

3.  •! 

[  a;-  —  y  =  5. 

[Suggestion.     First  solve  foi* 
x^  and  ^'^ ;  then  find  x  and  //.] 

|5A;2  +  3r2  =  32, 
*'    l2r2-3F  =  15. 

7(p  +  g)-8(p-g)=23, 
4(i)  +  9)+(p-9)  =  41. 

(c  +  a)  +  (a  +  ^)  =  1, 
6.     j  (a  +  6)  +  (6  +  c)  =  0, 

.  (6  +  c)  +  (c  +  a)  =  -  1. 

-  +  -  =  6, 
a;      y 


VIII:    CHAPTER  VI 

equations    for   the   unknown 


*    +^=3. 


5  —  x     y  +  1 
^     +^-=3. 


5  —  X*     2/4-1 
[Suggestion.     First  solve  for 


and 


10. 


3/  +  1-I 


«  ,   b 
X-     y 


X     y 
[Solve  for  o,  h ;  also  for  x,  y."] 


1   +  ' 


11.     ^ 


^     +-^  =  7 


1=4. 


1 

,aj      2/ 
[Suggestion 


First  solve  for 


y  -\-z     z-\-x 
[Suggestion.     First  solve  for 

,    and    — — ;    then 


-  and  -;  then  find  x  and  y.'] 


y 


X  +  y      X  +  z  7/4 

find  X  -\-  y,  X  +  z,  and  y  +  z;  then 

solve  for  x,  y,  z.'\ 


l  +  i-5 

V        W       O 

w      u      12 
n     V     ^ 


'     +^=5, 


12.     <^ 


X -y      x+y 

15 2_ 

,x  —  y     x-{-y 


[Suggestion 

1    1    1 

—  >  ~  >  ~ 

U      V     u 


First  solve  for 
;  then  find  n,  v,  ?<;.] 


ri 

a;2      ?/^ 


13. 


^  =  5, 


^  +  ^-  =  22. 
5  x^     y'^ 


13] 


REVIEW 


179 


14. 


x'^+y 


:.+ 


SX--27J- 

6 

cr  +  y-  '  3x^-2y- 


-.+ 


9 

2 


15. 


•  + 


=  b. 


m  —  >i      m  +  '*- 
[Solve  for  7n,  »?.] 

16.  A  and  B  can  do  a  piece  of  work  in  2  days ;  but  an  equal 
piece  of  work  wheu  A  puts  in  only  half  his  time  and  B  only 
one  third  his  time  requires  4^  days.  How  long  would  the  work 
take  A  and  B  each,  working  alone  ? 

17.  Two  pipes  empty  a  reservoir  in  1  hour  and  12  minutes. 
When  one  pipe  is  used  to  fill  and  one  to  empty  the  reservoir, 
it  becojnes  full  in  3  hours.  How  long  would  it  take  each  pipe 
separately  to  empty  the  reservoir? 

18.  A  boat  can  be  rowed  downstream  4  miles  in  the  same 
time  as  it  can  be  rowed  upstream  3  miles.  A  trip  6  miles 
downstream  and  back  requires  3|  hours.  Find  the  rate  of 
the  boat  in  still  water,  and  the  rate  of  the  current. 

19.  An  investment  yielding  simple  interest  amounts  to  S  2080 
in  4  years,  and  to  $2210  in  6  years.  Find  the  amount  of  the 
investment  and  the  rate  of  interest. 

20.  A,  B,  and  C  are  at  various  times  engaged  to  do  certain 
equal  pieces  of  work.  B  and  C  together  complete  the  work  in 
2  hours ;  C  and  A  in  li  hours ;  A  and  B  in  1-^  hours.  In  what 
tinie  can  each  do  the  work  alone,  and  in  what  time  can  all 
three  do  the  work  ? 


180  SIMULTANEOUS   LINEAR   EQUATIONS 


SUMMARY   OF   CHAPTER  VI:     SIMULTANEOUS  LINEAR 
EQUATIONS,  pp.  159-180 

Figure  :  two  intersecting  straight  lines. 

A  nsivers  :  pair  of  values  at  point  of  intersection. 

Algebraic  Solution,  Method  I :  remove  one  letter  by  addition  or  sub- 
traction. 

Elimination  :  any  process  for  removing  one  letter. 

§  86,  pp.  159-163. 

Formal  Rule,  Method  I,  by  Addition  or  Subtraction  :  essentially, 
multiplication  of  both  sides  of  each  equation  by  the  coefficient 
of  one  letter  in  the  other;  Exercises  I.  §  87,  pp.  163-165. 

Impossible  Case :  figure,  parallel  lines ;  no  answers ;  contradictory 
conditions,  revealed  by  attempt  to  solve ;  Exercises  II. 

§  88,  pp.  165-168. 

Equivalent  Equations :  figure,  coincident  lines ;  answers,  one  pair 
of  solutions  given  by  any  point  on  the  line. 

{intersecting! 
parallel      >  lines 
coincident  J 
no  solution     >  ;  Exercises  III.  §  89,  pp.  168-169. 

many  solutions] 
Solution  bij  Substitution  (Method  II)  :   essentially,  solve  one  equa- 
tion for   one  letter   and   substitute   this  value  in  the  other ; 
Exercises  IV.  §  90,  pp.  170-171. 

Solution  by  Comparison  {Method  III)  :  essentially,  solve  each  equa- 
tion for  one  of  the  letters  and  equate  the  values  found ; 
Exercises  V.  §  91,  pp.  172-17,^. 

Thi-ee  Unknowns :  elimination  of  one  letter,  reduction  of  problem 
to  two  equations  in  two  unknowns ;  Exercises  VI. 

§  92,  pp.  173-175. 
English  Problems :  solution  of  typical  examples ;  Exercises  VII. 

§  93,  pp.  175-177. 
Review  Exercises  for  Chapter  VI:  Exercises  VIIL        pp.  178-179, 


CHAPTER  VII.     SIMPLE   POWERS  AND    ROOTS 
PART    I.     POWERS    AND   ROOTS    OF   NUMBERS 

94,  Introduction.  The  student  is  already  familiar  with 
ordinary  powers  and  roots  (§§  9-10,  pp.  8-9). 

Ky  definition,  x''^  x  ■  x  •  x  •••  x  (w  times)  is  called  the 
nth  power  of  a.     (See  p.  8.) 

The  process  of  raising  a  number  to  a  power  is  often 
called  involution. 

By  definition,  \i  h  =  a  •  a  •  a  ■■•  a  (n  times),  then  a  =  Vi, 
i.e.  the  nth  root  of  h.     (See  p.  9.) 

The  process  of  extracting  a  root  is  often  called  evolution. 

[Let  the  student  again  state  the  definitions  given  in  §  9,  pp.  8-9.] 

There  are  often  two  answers  for  a  root  of  a  number. 
Thus,  4  =  2-2;  hence,  2  is  one  answer  for  the  square  root 
of  4.  But  4  =  (—  2)  •  (-  2)  also  ;  hence,  -  2  is  also  an 
answer  for  the  square  root  of  4.  When  no  sign  is  written 
before  a  V  we  shall  understand  that  the  positive  answer  is 
intended,  and  we  shall  agree  to  distinguish  between  the  two 
roots  by  trriting,  for  example : 

+  2  =  V4,         -  2  =  -  V4, 

the  negative  answer  in  any  case  being  said  to  be  the  nega- 
tive square  root.  Similarly,  ■Va^=  +  a,  although  —  a  is 
also  one  square  root  of  a^. 

There  is  only  one  cube  root  of  any  number  among 
numbers  we  know  at  present.  Thus,  8  =  2  •  2  •  2,  hence 
2  =  ^8,  but  -2  does  not  equal  ^8,  for  (_2)(-2)(-2)  = 
-8.     In  fact,  -2  =  ^-8. 

181 


182 


SIMPLE   POWERS   AND   ROOTS 


[Ch.  VU 


A  negative  number  has  no  square  root  among  numbers 
we  now  know ;  for  the  square  of  any  number  we  know  at 
present  is  positive,  except  zero,  whose  square  is  zero. 

In  fact,  since  an  even  number  of  either  +  or  —  factors 
gives  +  in  the  product,  an  even  root  of  a  negative  quantity 
is  meaningless,  at  present,  to  the  student. 

Such  roots,  even  roots  of  negative  quantities,  are  called  imaginary. 
Later  we  shall  discuss  them  (Appendix,  §  31),  and  we  find  it  possi- 
ble to  use  them  sometimes.  Just  now  we  shall  have  nothing  to  do 
with  them.  In  this  chapter,  in  even  roots,  we  shall  assume  that 
the  letters  used  have  positive  values. 

If  any  even  root  (i.e.  square  root,  fourth  root,  etc.)  of 
any  number  is  known,  then  the  negative  of  the  known 
answer  is  also  an  answer,  as  above. 

On  the  other  hand,  the  negative  of  a  known  answer  for 
an  odd  root  (i.e.  cube  root,  fifth  root,  etc.)  is  7iot  an  answer, 
for  the  odd  number  of  negative  signs  produce  — ,  not  +, 
in  the  answer.  There  is  only  one  answer  (among  num- 
bers we  know  at  present)  for  any  odd  root  of  a  number, 
since  an  increase  or  decrease  in  the  answer  would  cor- 
respond to  a  larger  or  a  smaller  number  than  the  given 
number. 

95.  Figure.  We  have  seen  how  to  indicate  the  squares 
of  numbers  in  a  figure.  (See  p.  30.)  Thus,  if  a:  =  a  giveh 
number  and  y  =  the  square  of  the  given  number,  we  have 
y  =  7?.  Let  us  give  x  various  values,  as  on  p.  80,  and  find 
values  of  y.     This  gives  a  table  as  follows : 


X 

y 

1 

1 

-1 

1 

2 
4 

-2 
4 

±3 
9 

±4 
16 

±5 
25 

±6 

&c. 

±1 

1 
5 

+  1 

4 
9 

±1 

9 
T 

±1 

&c. 

[Let  the  student  fill  in  the  blank  spaces  and  extend  the  table,  espe- 
cially by  putting  in  more  fractional  vahies.] 


94-95] 


POWERS   AND   ROOTS   OF   NUMBERS 


183 


Drawing  these  values  in  a  figure,  as  on  p.  30,  we  get  a 
graph  like  Fig.  ol.  Since  this  is  very  narrow  for  use, 
we  take  a  longer  unit  space  (one  heavy- 
marked  space)  horizontally  to  produce 
Fig.  32,  which  represents  the  same 
figure  on  an  enlarged  scale  horizontally 
(enlarged  one  way  only). 

To  find  the  square  root  of  a  number, 
we  merely  look  for  the  given  number  in 
the  table  marked  y ;  its  square  root  is 
the  corresponding  x,  for  we  squared  each 
X  to  get  the  corresponding  y. 

Or,  in  the  figure,  we  count  off  the 
given  number  vertically  (on  the  y  line), 
find  the  corresponding  point  on  the 
curve,  and  then  count  horizontally  to 
find  X. 

Thus,  if  2/  =  9,  we  rise  9  spaces  on  the  y 
line  ;  the  corresponding  point  of  the  curve  is  C ;  the  horizon- 
tal space  out  to  C  is  3.  The  point  C  is  also  at  a  height  9  ; 
its  a;  is  —3.     Hence,  the  square  root  of  9  is  either  3  or  —  3. 


Fig.  31. 


i 


-20- 


yMx^ 


B5< 


■10- 


Iho 


riSontalTij 


Fig.  32. 


184  SIMPLE   POWERS   AND   ROOTS  [Ch.  VII 

96.  Inexact  Roots.  There  are  numbers  that  are  not  in 
the  y  table.  For  example,  15  is  not  to  be  found  in  the  y 
table,  no  matter  how  complete  the  table  be  made. 

We  may  still  carry  out  the  process  in  the  figure.  Thus, 
rising  lo  points,  we  find  a  point  marked  Mow  the  curve; 
counting  horizontally  out  to  it,  we  find  x  =  3.8"^. 

In  fact,  we  have :   (3.8)2=  14.44,  ^nd  (3.9)2=  15.21. 

Hence,  (3.9)^  is  too  large,  while  (3.8)^  is  too  small; 
that  is,  a:-  =  3.8  gives  a  point  {x  =  3.8,  y  =  14.44)  below  M, 
while  a^  =  3.9  gives  a  point  {x=  3.9,  y  =  15.21)  above  M. 

With  a  more  precise  figure  we  could  find  values  of  x  still  closer 
together -J— differing  in  any  place  of  decimals  we  please  —  which  give 
points  above  and  below  M.  Thus,  while  many  numbers  do  not  give 
exact  square  roots,  we  can  interpolate  the  number  between  two  which 
differ  in  a  small  amount.  This  process  gives  a  result  to  any  number 
of  decimals  desired,  but  it  is  not  exact.  The  negative  of  the  answer 
thus  found  is  also  an  answer,  as  above. 

Any  inexact  root  may  be  left  merely  indicated  by  means 
of  the  radical  sign ;  thus  Vl5.  Any  expressed  root, 
whether  exact  or  inexact,  is  called  a  radical.  Any  ex- 
pression which  contains  a  radical  is  called  a  radical  expres- 
sion.    (See  also  pp.  186,  192,  284.) 

97.  Formal  Process.  A  process  for  finding  the  decimal 
places  is  based  on  the  rule  (x  +  yY'  =  x^  +  2  xy  -\-  y^,  p.  93. 

We  found  VT5  =  3.8+.  Suppose  the  digit  in  the  next  place  of 
decimals  is  d,  then  squaring  -3.8  +  f/(.01),  we  find 

{3.8  +  (/(.01)}2  =(3.8)2  +  2,/ (3.8)  (.01)  +  r/2(.()001) 

=  (3.8)2  ^  d{.07Q)±  (very  small  number). 

Since  we  wish  to  get  near  to  15,  we  write 

(3.8)2+  ,/(.076)=  15  (nearly), 

or,  d(.07Q)  =  15  -  (3.8)2  (nearly), 

,      15 -(3.8)2.         ,\ 

or,  ^  rf  = i — ^  (nearly). 

.076 

We  can  tell  by  this  very  closely  what  d  must  be ;  since  d  is  an  integer, 

we  can  guess  it  exacth/. 


fJ(i-97] 


TOWERS   AND    ROOTS   OF   NUMBERS 


185 


The  work  is  written  as  follows 


15 
14.44 


18.8  17 


2  X  3.8  X  .01  =  .0760 

(.01)-  X  7  =  .0007 

.0767 

We  continue  upon  the  same  plan : 

2  X  8.87  X  .001      =  .007740 
==  .000002 


..5600 
.5869 


(.001)5 


.007742 


2  X  3.872  X  .0001  =  .00077440 
(.0001)-  X  9  =  .00000009 


.00077449 


.0281 

.023100 
.015484 


18.871219 


.00761600 
.00697041 


.00064559 

and  so  on.  We  really  omit  many  of  the  decimal  points  and  zeros, 
these  are  useless  if  we  keep  the  places  rif/ht ;  notice  that  in  order  to  do 
so  two  zeros  are  to  be  placed  after  the  difference  after  subtraction, 
and  one  zero  after  the  ^^  trial  divisor."' 

The  trial  divisor  is  twice  the  amount  previously  found  with  the 
decimal  point  shifted  as  above. 

The  whole  work  in  the  above  example  would  be  written  as  follows : 

15.00000006 1 8.8729,  etc. 


9 

2  X  8  =  60 

600 

8 
68 

544 

2  X  88  =  760 

5600 

767 

5369 

2  X  887  =^  7740 

28100 

2 

7742 

15484 

2  X  3872  =  77440 

761600 

9 

77449 

697041 

64559,  etc. 

(The  small  dots  placed  over  every  other  place  beginning  with  the 
units'  place  in  the  original  number  are  helpful  in  bringing  down  the 
figures,  especially  if  the  given  number  contains  decimal  digits.) 


186  SIMPLE   POWERS   AND   ROOTS  [Ch.  VII 

The  result  obtained  is  a  decimal  fraction,  whose  square 
is  less  than  15 ;  but  the  square  of  the  decimal  just  higher 
in  the  last  place  is  greater  than  15,  at  each  stage : 

(3.8)2<  15  <  (3.9)2, 

(3. 87)2  <  15  <  (3. 88)2, 

(3.872)2<15<(3.873)2. 

We  say  that  Vl5  is  determined  by  this  sequence  of 
numbers,  i.e.  by  the  decimal  fractions  found  as  above. 

(This  is  really  the  first  definition  of  VT5;  the  idea  of  drawing 
a  smooth  curve  through  the  points  we  can  locate  in  Fig.  31  contains 
the  essentials  of  a  strict  definition.     See  A2:)pendix,  §  29.) 

In  carrying  out  any  operation  upon  Vlo  we  shall  really 
operate  upon  these  decimal  fractions;  the  result  is  said 
to  be  the  number  defined  by  these  decimal  results.  (See 
also  Appendix,  §§  29-30.)  The  same  statements  apply,  of 
course,  to  all  inexact  square  roots  as  well  as  to  V15. 

Any  inexact  root,  i.e.  a  root  that  cannot  be  expressed 
exactly  in  fractions  or  integers,  is  called  a  surd.  Any 
expression  containing  a  surd  is  called  a  surd  expression. 
The  quotient  formed  by  dividing  any  integer  by  another 
intesrer  is  called  a  rational  fraction.  A  number  that  is 
an  integer  or  a  rational  fraction  is  often  called  a  rational 
number.  Any  number  that  is  neither  an  integer  nor  a 
rational  fraction  is  called  irrational.  (See  pp.  192,  284, 
and  Appendix,  §  29.)  Thus,  every  surd  is  an  irrational 
number. 

EXERCISES   I  :    CHAPTER   VII 

¥ind  the  square  roots  of  the  following  numbers: 
•     1.   441.  4.    1253.16.  7.    .0625. 

2.  324.  5.    24.1081.  8.    .1369. 

3.  55,225.  6.   27.6676.  9.   .056644. 


97-98] 


POWERS   AND   ROOTS   OK   NUiMBERS 


187 


From  the  figure  on  p.  183  (y  =  .t-'),  read  oft'  the  values  of  the 
following  radicals : 


10.    V2.       11.    V6. 


12. 


10. 


13.    Vl8.       14.    V2.5. 


Calculate  the  values  of  the  following  radical  expressions  to 
three  places  of  decimals : 


15.  V2. 

16.  VO. 


17.  Vl2. 

18.  V33. 


23.    \2-V3. 


19.     V19. 

20.  VtT. 

24.    Vt  -  V  Vl7^ 


21.  V109. 

22.  V2  +  V3. 


98.    Powers  of  Radicals.     We  are  justified  in  saying  that 
the  square  of  V2  is  2,  or,  in  general,*  if  a  is  positive, 

If  we  do  so,  we  can  easily  square  such  expressions  as 
3+2V5. 

For  by  p.  93,  Chapter  IV,  we  have 

hence,  (3  +  2  V5)2  =  (3y  +  2(3)(2\/5)  +  (2  V^)2 

=:9  +  I2V5  +  4(\/5)2 
=  9  +  12\/5  +  20 
=  29  +  12  Vs. 

Check:  \/5  =  2.2-36  (nearly). 

Now  3  +  2V5  =  7.472  (nearly), 

and  29  +  12\/5  =  55.832  (nearly), 

,  (7.472)2  =  .55.832  (nearly). 

'  We  shall  need  this  process  in  the  following  chapter. 
Other  operations  of  this  kind  will  require  some  further 
explanation  and  are  deferred  until  Chapter  XI,  p.  284. 

t  *  See,  however,  Chapter  XI,  p.  280,  and  Appendix,  §  31. 


188  SIMPLE   POWERS   AND   ROOTS  [Ch.  VII 

EXERCISES  II:  CHAPTER  VII 

Square  the  radical  expressions  : 

1.  2  + Vs.  6.   2V7-5.  11.    V17-4. 

2.  2-V3.  7.    1V3-1.  12.   2VTi-7. 

3.  1  +  3  V2.  8.    a;  +  Vy.  13.    a  V&  -  c. 


4.  SV'*!  — 7.  9.    a;+Va— a?-.     14.   mx-\-m^y'^—oi?. 

5.  V8-3.  10.  1-V^^^.      15.   kx-Wf  +  1. 

16.    kx  +  lyjf-^- 

99.  Simple  Operations.  We  shall  also  need  to  notice 
that  Vl2=  V4T3  =  V4V8=  2V3  for  example;  or,  in 
general,  Va%  =  Va^Vb  =  aV3,  if  a  and  b  are  positive, 
This  is  justified  by  observing  that 

(2V3)2  =  4(V3)2=12,    whence  2V3=VT2; 
or,  in  general,  if  a  and  b  are  positive 

(aVly  =  a\^by  =  a%,    whence  a V^  =  Va^J. 

T-i       .  ^/3      V3      V3 

Likewise,  v- =  —  =  — ; 

or,  in  general,  -y—  =  ~—  = 

These  operations  are  more  fully  studied  in  Chapter  XI, 

p.  284. 

EXERCISES  III:  CHAPTER  VII 

Remove  from  under  the  radical  sign  all  square  factors  in 
the  following,  assuming  that  the  letters  used  have  positive 
values: 

1.    Vl8.  5.    V98.  9.    VT2]V.      13     ^S". 


2.    V50.  6.    V24.  10.    Vf^g.  ^^^ 


3.    Va^d'.  7.    V99.  11.    Vf 


14. 


4.    V75.  8.    -^d^xf.  12.    VX.  ^dV/ 


98^100]         POWERS   AND   HOOTS   OF   NUMBERS 


189 


a/!(= 


'x>/ 


16.    Va  (=  VfW). 


17.    V^  (=  VtVir)- 


18.     V|. 


Express  the  following  quantities  in  a  form  entirely  under 
the  radical  siirn  : 


19.    5V3. 

22. 

7)in^ii2). 

25. 

?V6c. 

0 

20.    3V5. 

23. 

iV3. 

26. 

m  In  ^ 
11   *  ))i 

21.    6V3. 

24. 

tVV8. 

27. 

a-v 

a-'+y 


ar-2/- 


100.    Cube   Roots. 

analogous  process. 


We    may    find    cube    roots    by    an 
First  let  us  draw  a  figure,  where  x 
means  any  number  and  i/  means  the  cube  of  that  number 
i.e.  y  =iy?.      A  table  is 


X 

V 

0 

0 

+  1 
+  1 

-1 
-1 

+  2 

+8 

-2 
-8 

±3 

±27 

±4 
±6i 

±5 

±6 

etc. 

±1 

±f 

±¥ 

±1 

±1 

etc. 

[  Let  the  student  fill  in  the  blank  spaces  and  expand  the  table.] 

From  this  table  we  may  draw 
a  figure  that  will  appear  as 
shown,  the  scale  being  taken  for 
convenience : 

1  vertically  =  1  small  space ; 
1  horizontally  =  5  small  spaces. 

From  this  figure  we  can  read 
off,  approximately,  the  cube  of 
any  number,  or  also  the  cube 
root  of  any  number,  as  for  square 
roots  above. 

Thus,  the  cube  of  1.8  is  seen 
to  be  about  of  (really  5.832); 
the  cube  root  of  11  is  seen  to  be 
a  little  over  2.2.  INIore  accurate 
results  can  be  found  by  drawing 
the  figure  on  a  larger  scale.  Fig.  33. 


t:    : 

:: — 

/- 

_            -  f-  - 

-Tt- 

:"    :      '\  -  - 

: :_;::..] 

.  ..  X 

1  r  i  r. ;  y :  -  - 

:  ^"Z""-\"T-'-'-""-- 

..-.-..^ 

_  _  .z.      

z  _.  

___..^..5......                        .....^^ 

_,-.          . -_          ^ 

?_::...:.:__.;(.:-:*:--: 

-f-            -,  f-^ixH    }erru2Uyr- 

190 


SIMPLE   POWERS   AND   ROOTS 


[Ch.  VII 


A  method  similar  to  that  of  §  97  can  be  devised  for  find- 
ing further  decimal  places.  (See  Appendix,  §  20.)  But 
it  is  really  seldom  used  because  there  is  a  much  quicker 
process.      (See  Chapter  XIV  on  Logarithms,  p.  339.) 

101.  Higher  Roots.  A  graphical  process  like  that  above 
applies  to  finding  any  root  of  any  number  approximately. 
The  figure  shows  the  pictures  for 


—  ^i 


and 


0)    ^  = 

(2)     ij  =  x^, 
Y 


:::::::--::::--:  -"t:::: 

_    _         -       jt 

3 

_     _            .          I    

._     I 

:              -         _ .    JL  _  _  _ . 

-      lT-      -- 

J       

J    - 

30 

1 

::::::::::::]:=:-::::: 

" i_  I 

:  --sdt     - 

.    -h?  -  -  sii-            i  - 

.      H^.    -22                --t- 

S    a' 

20 

— Jit,, 1-- 

::::::::::::S=  ===;=?  =  =  = 

X              -tf  -  -    ^/ 

:          df-i--  ?k- 

JL   [     ^i^ 

/  J       / 

-  -  i ^If 

f  J     /       -^ 

k 

:<=. 

'  ^S 

'   ^~X'  ■   //  y  --I- 

^^         yvo  ■     ,     .  it             _    - 

^--.  >).  ..^ ■  ■  ■  0   ■     •  ■ 

^^      ■    -  ^             : 

'"T'  -*^                            '  i 

/I     '»/'''                                                               V    .'<  n  l' 

kI                                          Viui 

i\'  certi\-'iliij  f- 

I                              -tin 

X 


Fig.  34. 

from  which  fourth  and  fifth  roots  may  be  found  approxi- 
mately.    These  roots  may  be  found  more  accurately  either 

(1)  By  drawing  a  larger  figure. 

(2)  By  a  method  similar  to  that  of  §  97. 

(3)  By  logarithms.     (See  Chapter  XIV,  p.  339.) 


100-101]  roWEHS   AND   ROUTS   OF   NUMBERS  191 

'J'lie  last  method  is  by  far  the  easiest;  further  explana- 
tion of  it  will  be  found  in  the  Chapter  on  Logarithms 
(Chapter  XIV). 

EXERCISES   IV:    CHAPTER   VII 

Draw  accurately  figures  for  y  =  x-,  y  =  x*,  y  =  ^,  as  indi- 
cated on  p.  190,  and  estimate  as  accurately  as  possible  the 
values  of  the  following  radicals;  check  by  raising  the  result 
to  the  proper  power: 

1.  ^^.        3.   Vri.         5.  ^50.         7.   ^^50.         9.  -^76. 


2.  </17.        4.  ^25.         6.  a/31.         8.  ^60.       10.  A/'IOO. 

11.  How  many  cube  roots  has  a  positive  number  among  the 
numbers  we  already  know  ?  a  negative  number  ? 

12.  How  many  fourth  roots  has  a  positive  number?  a  nega- 
tive number  ? 

13.  How  many  fifth  roots  has  a  positive  number  ?  a  nega- 
tive number? 

14.  If  .?/  =  .^'^  what  is  x  in  terms  of  ?/?  By  noting  corre- 
sponding values  of  x  and  y  in  Fig.  33  on  p.  189,  and  plotting 
them  reversed,  draw  the  figure  for  y  =  '\/x. 

15.  Draw  the  figure  for  y  =  -y/x. 


PART    II.     POWERS    AND   ROOTS    OF   MONOMIALS 

102.  Notation.  Instead  of  writing  V~,  V  ,  ^  ,  •••,  etc., 
for  roots,  we  may  write  a  small  figure  |  or  |^  or  |^  or  ^  •••, 
etc.,  at  the  upper  right-hand  corner,  in  the  place  usually 
occupied  by  an  exponent. 

V4  =  4^=2,    -V4=-4^  =  -2, 

^8  =  8*  =  2,    ^/'^8  =  (-8)^  =  -2. 
_       1 
So,  in  general,      Va  =  a". 

This  notation  is  very  convenient ;  it  does  not  cause  any 
confusion,  for  we  have  never  written  fractions  in  the  ex- 
[)onent  position  before  this ;  hence,  there  is  nothing  to  be 
understood  by  4^,  for  example,  except  V4. 

Any  expression  that  contains  a  root  sign,  either  by 
use  of  the  symbol  V  or  by  means  of  a  fraction  in  the 
exponent  position,  is  called  a  radical  expression. 

103.  Monomials.  Monomials  may  be  raised  to  powers, 
or  their  roots  may  be  taken,  according  to  the  rules  for 
multiplication. 

Ex.  1.    (4  ahy  =  (4  ab-)(A  ab-)(4:  aV)  =  64  a%\     (See  p.  73.) 


Ex.  2.    16  a'h~  =  4  a^d  •  4  w^ft ;  hence  Vl6  a*¥  =  4  a-h. 


Notice  also  -  Vl6  a^"^  =  -  4  a%. 

In  raising  to  powers  or  in  taking  roots,  care  must  he  taken 
with  regard  to  signs.      (See  Rule  of  Signs,  p.  71.) 

•p,      o     /^  ~  2  xy"^-  __  f  —2  xy^  (  —2  'X]f\  _  +  4  '3?]f' 


3  a     J      \    3  a    J\    3  a     J         9a^ 
192 


POWERS   AND    ROOTS   OF   MONOMIALS  193 

—  27  «''x'"       f—3  ax^\  /  —  3  ax-\  /  —  3  oa:"-^ 


Ex  4 

8r  V    22/    A     2y    A    2y 

hence,       'j- 27  aF^  ^- 3ax\    /     ^  ±3aar\_ 
\       82/«  2y      '    I,  2y     J 

A  simple  power  (p.  8)  is  one  wliose  degree  is  a  positive 
integer,  such  as  2,  3,  etc.     Thus,a3  is  a  simple  power. 

EXERCISES    V:    CHAPTER   VII 

Perform  the  indicated  involutions  and  evolutions: 

1.  (a'bf.  7.  v^yv^.  12.  (-^;':^y 

2.    (Sxyzy.  8.    ^^lm*n'r 


ps-v 
xy 


10.    V-32a;'«.  ^*-    \-^^^.i:^ 


6^0^ 


5.  (3a;«2/W-  I - 

11     /_^Y  15       '   -  — 

6.  ^-64a'^6«.  ■   \2hcyj'  '     ^        a^aj^*^ 

104.  Formal  Rule,  Monomials.  We  may  make  a  rule 
by  the  reasoning  above.  Thus,  if  m  and  n  denote  two 
positive  integers, 

I.  (x"y"  =  x"  •  a;"  •••  x"(m  times)  =  a;""'".    (See  rule,  p.  72.) 

To  raise  a  letter  with  an  exponent  to  a  simple  poiver,  mul- 
tiply the  exponent  by  the  degree  of  the  power.  In  short,  in 
involution  multiply  exponents. 

It  is  well  to  contrast  this  strongly  with  the  rule  for 
multiplying  powers,  p.  73,  which  is 

II.  x"  X  a:'"  =  x«+"'. 

In  multiplying,  add  exponents. 
hedrick's  el.  alg. — 13 


194  SIMPLE    POWERS   AND   ROOTS  [Ch.  VI] 

We  also  have 

III.    (xyy^  =  x  •  y  ■  X-  y  •  X  •  y  ••'  X  •  y(n  times) 

=  [x  '  X  ■  x---x(ri  times)]  x[y  •  y  ■  y  •••  y(n  times)]  —x"  x  ?/". 

Any  simple  power  of  a  product  is  equal  to  the  product  of 
the  same  powers  of  the  separate  factors. 


Ilia.  Similarly,  — =  -  ]  •  This  rule  follows  from 
III,  also:  ^"      ^^^ 

Any  simple  power  of  a  quotient  is  equal  to  the  quotient  of 
the  same  poivers  of  numerator  and  denominator. 

These  rules  assist  in  reducing  the  work  of  involution: 
To  raise  a  monomial  to  a  simple  power.,  raise  the  coeffcient 
to  that  power,  multiply  the  exponent  of  each  letter  hy  the 
degree  of  the  power,  and  make  sure  that  the  siynfoUotvs  the 
Rule  of  Signs. 

The  Rule  of  Signs  (p.  71)  shows  that 

(1)  an  even  number  of  negative  factors  gives  +  in  the  product; 

(2)  an  odd  number  of  negative  factors  gives  —  in  the  product; 

(3)  any  number  of  positive  factoi-s  gives  +  in  the  product. 

This  rule  is  simply  the  written  expression  of  what  we 
already  did  in  §  103  ;  if  this  rule  is  forgotten,  the  examples 
can  all  he  solved  hy  §  103. 

105.  Advantage  of  Fractional  Notation.  Practice  with 
the  fractional  notation  of  §  102  shows  its  convenience. 


Ex.  1.     V16  a'h''  =  4  aH),  for  (4  n'h)  (4  n'h)  =  16  a'b\ 
or,  (16a*&-)*  =  4a-/y. 

Note  that  this  follows  the  Rules  of  §  104 ;   for  the  work 
(16  a%'^y  =  {\&)^i^''^l?'^^  =  4  a'h 
is  correct,  although  we  had  no  reason  to  expect  it. 


lU4-10a] 


J'OWKKs   AM)    KUUTS   OF   MONOMIALS       195 


Ex.  2.  Trying  the  same  rules  of  §  103  for  A/-27a'V,  Ave 
find: 

(-  27  a'j^)^=(-  27)V^  3/xi^  _  3  ^^.2^ 

?'•// /c/*  /s  correct,  for  (  —  3  aor)  (  —  3  ax'-)  (  —  3  aor)  =  —  27  a'V'. 

It  is  necessary  for  the  student  to  check  his  work  as 
above,  to  avoid  errors,  at  least  until  he  becomes  expert. 

Rule  I  of  §  104  holds  when  the  power  is  a  fraction, 
that  is,  for  roots,  if  the  root  can  be  found  otherwise  (i.e. 
for  exact  roots.     See  §  94,  p.  182,  and  §  98,  p.  187): 


,t•«^„  —  f^^-JJ-re 


(a'-"y 


for 


aJ^  ■  aJ'  '••  a''{;ri  times)  =  a; 


—  nk'n 


Later  (§  136,  p.  285),  we  shall  see  that  the  rules  of  §  104  always 
hold  when  we  properly  extend  our  notation. 

Likewise,  Rule  III  holds  for  exact  roots : 
1  1  1 

(a")"  X  (5")"  =  (a»i")",  for  each  side  is  equal  to  ah. 


EXERCISES    VI:    CHAPTER   VII 

Perform  the  following  involutions  by  the  aid  of  Rule  I : 
-2cehcy.  7. 

xy^y.  3 

5  mn-jfy.  9. 

[  —  ax'y-'z^y. 


1. 
2. 
3. 
4. 
5. 
6. 


10. 


{x"'-"y"-Pz''-^y. 

f-2x-fzy^ 

VI 1  a'^z"-^  J  ' 

/-  5  mVpy 
V    3  r^'s"^   J  ' 


Perform   the  following  evolutions   by  the  aid  of   Rule  I; 
check  by  actual  multiplication,  when  the  index  is  numerical : 


11.  V64a*6«. 

12.  -v/-64x-«?/V'. 


13.  y/21S7  p'q^K 

14.  </6257s^. 


196  SIMPLE   POWERS    AND   ROOTS  [Ch.  VII 


15.    V-27z«.  18.    Va^^ci*. 


106.  Longer  Expressions.  Binomials  and  longer  ex- 
pressions can  be  raised  to  powers  by  simply  applying 
the  rules  for  multiplication ;  or  also  by  using  the  rules 
of  Part  III  of  Chapter  IV,  p.  91. 

(a  ±hy^a^±2ab  +  h"-.      (See  p.  93.) 

(a  ±  by  =  rt3  _t  3  a%  +  3  ah"^  ±  h^.      (See  p.  99.) 

(a  ±hy  =  a^±^  a%  +  6  aW  ±^a¥  +  bK      (See  p.  99.) 

Ex.  1. 

(4  xY-  -  3  xfY  =  (4  xh/y  -  3(4  .r''?/2)2(3  a;/)  +  3(4  a^y^)(3  xff 
-  (3  xff  =  64  .-«»?/« -  144  xhf  + 108  x^?/«  -  27  ary. 

Test  this  by  direct  multiplication  of  three  factors,  each 
4  3^y^  —  3  xy^. 

Roots  of  longer  expressions  may  be  found  by  ins^jec- 
tion,  comparing  with  the  formula  written  above. 

Ex.  2.    Find  the  square  root  of  4  m~  —  12  mn  +  9  n^. 

Comparing  with  (a  —  b)'^  we  see  o  =  2  m,  b  —  S  7i  will  fit  to  give  the 
given  expression  : 

(2  m  -3,1)^  =  4  m^  -  12  mn  +  9  n^ ; 

hence,  (4  m^  —  12  mn  +  9  n'^)-  =  (2  ??j  —  3  n). 

Another  answer  is  —  (2  m  —  3  n),  by  the  general  rule  that  the  negative 
of  one  answer  for  a  square  root  is  also  an  answer ;  for, 

[_  (o  „;  _  3  „)-]2  =  (_  2  „,  +  3  „)2  =  4  7^2  -  12  mn  +  0  n^. 

It  is  not  certain  which  of  these  answers  is  the  negative  one  until 
we  know  whether  2  7n  —  3  n  is  positive  or  negative.  If  m  =  5  and 
7)  =  1,  2  ?«  —  3  n  =  7 ;  (—  2  m  +  3  n)  =  —  7  ;  the  given  expression  is 
of  course  the  square  of  this,  i.e.  49. 


105-107]       I'UWEK.S    A'SD    ItUOTS   OF    MONOMIALS         197 

Check:  If  »h  =  5  and  n  =  1, 
im^-12nin  +  [)»-^  =  -i{oy-  12(5)  (1)  +  9(1)2  =  100-60  +  9  =  49. 
A  numerical  check  uf  tliis  kind  is  most  valuable  in  avoiding  errors. 

Ex.  3.    Find  the  square  root  of 

4  or  +  y  //-'  +  IG  z-  -  12  xi/  -+■  16  xz  -  24  yz. 

Comparing  with  a-  +  //-  +  c-  +  2  ah  +  2  ac  +  2  he  =  (a  +  ?>  +  c)^, 
from  p.  92,  vre  see  that  a  =  2  u,  /y  =  —  3  ^,  c  =  4 ;:  will  fit ;  for, 

(2  a-  -  3  7/  +  4  3)2  =  4  x2  +  9  //^  +  16  s^  _  lo  x^  +  16  x^  -  24  yz. 

Hence,  (4  x^  +  9  j/^  +  16  ^2  _  12  ^.^  ^  jg  ^.^  -  24  ^z)  -'  =  2  x  -  3  ?/  +  4  «. 

C^eci-  .-Put  X  =  1,  ^  =  2,  3  =  3. 

[4(l)2  +  9(2)2+16(3)2-12(l)(2)  +  16(1)(3)  -  24(2)(3)]?  =  64i  =  8. 

2  X  -  3  ?/  +  4^  =  2(1)  -  3(2)  +  4(3)  =  8. 
Another  answer  is  -  (2  x  -  3  ^  +  4  ;)  =  —  2  x  +  3  3/  —  4  2.     Check  it. 

EXERCISES    VII:     CHAPTER    VII 

Perform  the  indicated  operations  : 

1.  {ax  +  hyf.  3.    (2;>-7(/)'.  5.    {111  + In)*. 

2.  {ab-.vy)-.  4.    {Sa-2by.  6.    (c-2cl  +  e)-. 

7.  (bc  +  ca  +  ab)-.  12.    («-"'  + 12  trt"  +  3G  ^-'')'. 

8.  (2  .f -3^  +  4  2)2.  /       o„ 

9.  (ap-bqy+{ab+pqy.  \        ot 

10.  (..^-8.i-='+24;r-32a-+16)l     ^^    (^I^  +  vt)'. 

11.  (2'^  -  12  ^2  _^  48  2  -  64)i  ^^'         / 

15.    (4  a;-  +  y-  +  9  2;-  -  G  ?/2  +  4  xy  -  12  2!.f)i 

107.  Process  for  Square  Roots.  Roots  of  longer  expres- 
sions may  also  be  found  by  the  following  process,  which 
is  occasionally  convenient.  In  simple  examples,  however, 
roots  can  be  best  found  by  inspection  when  there  is  any 
exact  answer. 


198 

We  know 


SIMPLE    POWERS   AND    ROOTS 


[Ch.  Vll 


(a  +  6)2  =  a2  +  2  ah  -|-  i^ 
=  a2  _|-  (2  a  +  h)h. 

This  we  write  as  follows : 

a^  +  'lab  +  lPAa  +  h 


a2 

2a 

+  h 
2a+b 

2  «6  +  h'^ 
lah  +  W' 

If  there  were  more  terms,  we  should  proceed  similarly ; 
the  important  step  is  to  form  the  "  trial  divisor "  (2  a, 
above),  by  multiplying  the  part  already  found  by  2  ;  to 
this  we  add  the  second  term  (6,  above)  and  multiply  the 
sum  (2  a  +  6,  above)  by  this  second  term  ;  this  gives  the 
total  original  amount  as  shown  above. 

Ex.  1.  Find  (4  m-  -  12  mn  +  9  m-)^  (see  Ex.  2,  §  106,  p.  196). 
The  work  may  be  written. 

4  m^  -  12  mn  +  9  n^\2  m  -  3n 
4  m^ 
2  X  2  7«  =  4  ?n 


-3n 


4  m  —  3  n 


—  12  mn  +  9  n^ 

-  12  mn  +  9n^ 


Ex.  2.    Find  (4  x"  +  9  /  + 16  z^  _  12  x-i/  + 16  xz  -  24  ?/2)  K 

4  x2  +  9  //H-  16  2^  _  12  xi/  +  IGxz-  24 y::|2x  -  :i  y  +  4.; 
4  a;'-^ 


2  X  2  a:  =  4  a; 

-3?/ 

9  (/2  +  16  ^2  _  12  xy  +  16  xz  -  24  .yz 
(arranged) 
-  12  xy  +  9  ?/2  +  16  2^  +  16  xz  -  24  yz 

4  a:  —  8  ?/ 

-  12  xy  +  9  //2 

2(2  X-  -  8  ?/)  = 

4x-62/ 

+  4z 

16z2+  16xz-24y/2 
(arranged) 

4x  —  6?/  +  42 

16X2-   24^2+    16  22 
16x2-24?/2+    16  22 

107]  POWERS   AND   ROOTS   OF   MONOMIALS  199 

In  this  example  the  terms  cannot  be  arranged  until  we 
see  which  term  will  afford  an  exact  quotient  with  the  trial 
divisor.     Practice  will  indicate  the  best  arrangements. 

Only  exact  square  roots  should  be  attempted  by  this 
process.  The  process  is  valuable  also  in  discovering 
whether  or  not  a  given  expression  is  a  perfect  square. 
Inexact  roots  are  treated  (for  numbers  only)  in  Part  I, 
Chapter  VII,  §  97,  pp.  184-18G. 

EXERCISES  VIII:  CHAPTER  VII 

Extract  the  square  root  of  the  following  expressions: 

1.  \&p^  +  2oq--^0i)q. 

2.  9  c-  +  400  iVx"  + 120  cdx. 

3.  a-  +  25  h-  +  9  r-  -  30  6c  +  6  ca  - 10  ah. 

4.  X*  -f  2  x^y  +  3  x'lr  +  2xf  +  y*. 

5.  Ax*-V2x'  +  \3x'-Qx  +  l. 

6.  4  —  4  a  —  7  a-  +  4  a^  -\-  4  a*. 

7.  9  x^  - 12  x^  -  26  x"  +  20  a-  +  25. 

8.  x^  —  2  x?y  —  :iry-  -\-  2  xy^  +  y*. 

9.  a^-2 a'  +  3  a'  -  4 a^  +  3a-  -  2  a  +  1. 

10.  l-o(66+4)+a-(9  6-  +  126  +  4). 

11.  9  a«  -  12  a'h  +  10  a'b^  -  4  d'U'  +  ci'bK 

12.  ^^-A2t  +  Slf-12f  +  ^t\ 

13.  4  6-c-  -  16  ahc'  +  16  a^c''  -  12  ab-c  +  24  d'bc  +  9  a^ft-. 

14.  a*  +  4  a^/j  +  6  a^fe^  _,_  4  ^,53  ^  ^4_ 

15.  r'^  -  6  ?-^  + 15  r's'  -  20  v^.s''  -(- 15  ?-V  -  6  j-s''  +  s". 


200  SIMPLE    POWERS   AND    IKXrrs  [Ch.  VII 


REVIEW  EXERCISES  IX:  CHAPTER  VII 

By  use  of  the  figures  on  preceding  pages,  estimate  the  fol- 
lowinsf  roots : 


1.  V60.      2.  {/60.     3.  -v/eo.      4.  v60.     5.  -xVyo. 

By  rules  for  use  of  fractional  exponents,  so  far  as  they  have 
been  given  (§  105,  pp.  194-195),  show  that 

6.  \V60  =  ^60.  8.    Vvc=  vc  =  VVc. 

3 _ _ 

7.  \Vx  =  ^x  =  \y/x.  9.    (Vxy  =  x. 

Plot  the  following  curves,  making  all  necessary  calculations 
by  use  of  the  figures  (Figs.  32,  34,  pp.  183,  190). 

10.  y  =  X  +  x^.  12.    y  =  .i'  +  -\/x.      14.    y  =  ■\^x^. 

11.  y  =  x  —  Va:.  13.    y  —  ^x\  15.    y  —  xr  +  Vx 

Perform  the  operations  indicated : 

16.  (5-2V6)-.  20.  (5V2  +  T)(5V2-7). 

17.  (5  +  2V6)l  21.  (3V^-7y/)-. 

18.  (5  +  2V6)(5-2V6).  22.  {Cx-y+hzf. 

19.  (5V2-7)2.  23.  {x  +  y  +  zy. 

24.  (27  cv"  -  27  a-  +  9  o  -  1)1 

25.  (16  X*-  96  x^y  +  216  xY-  216  x-/  +  81  /)l 

Find  the  square  root  of  each  of  the  following  expressions : 

26.  9x^''  +  6a;«-5x«-2.ir^  +  l. 

27.  x-"*  -  6  2^y  +  13  ar^?/-  —  12  xif  +  4  ?/\ 

28.  a2-2a?>-362  +  ^'  +  ^. 

a        (t- 

29.  a^  +  62  +  0^  +  rf^  +  2  rr/>  +  2  ax  +  2  ?>r^'  +  2  cxX'  -\-  2  a-rf'-'  +  2  &c. 

30.  .1*  +  2  ;7-^  +  3  a;^  +  4  ic^  _^  o^^i  +  2  .»•  +  1 . 

31.  a«  -  4  o^  +  10  a*  -  20  cv"  +  25  «-  -  24  a  +  16. 


107] 


SUMMARY  201 


SUMMARY  OF  CHAPTER  VII:    SIMPLE  POWERS  AND  ROOTS, 

pp.  181-20U 

PaKT  I.       POWEKS  AND  RooTS  OF  NuMBERS.  pp.  181-192. 

Definitions:  elemeatary  definitious  recalled  ;  involution  and  evolu- 
tion. 

Nutnber  of  Roots:  odd  roots,  one  answer;  even  roots  of  positives, 
two  answers;  even  roots  of  negatives  ("imaginary  numbers"), 
no  answers  among  numbers  now  known  to  student. 

Notation  :  V  sign  for  even  roots  denotes  the  positive  answer ;  the 
negative  answer  indicated  by  —  V".  §  91,  pp.  181-182. 

Figure  for  Square  Roots  :  graph  of  ^  =  x-  drawn  ;  square  roots  from 
figure.  §  9.5,  pp.  182-183. 

Inexact  Square  Roots  :  approximate  square  roots  of  inexact  squares 
found  from  figure.  §  96,  p.  184. 

Formal  Process  for  Square  Rants :  arithmetic  process  for  closer  ap- 
proximation ;  formal  process,  using  "  trial  divisor  " ;  root  actu- 
ally defined. 

Definitions:  rational  fraction  —  quotient  of  tw^o  integers ;  rational 
number  —  integer  or  rational  fraction  ;  irrational  number  — 
not  rational ;  surd  —  irrational  radical.     Exercises  T, 

§  97,  pp.  184-187. 

Squares  of  Radicals:  (Vay^  =  a;  longer  forms  by  previous  rules. 
Exercises  11.  §  98,  pp.  187-188. 

Siinple  Operations  on  Quadratic  Radicals :  insertion  and  removal  of 
simple    factors ;    simple  multiplications  and  divisions  by   in- 
tegers.    Exercises  III.  §  99,  pp.  188-189. 
Cube  Roots  :  figure,  y  =  x^;  approximate  cube  roots  ;  other  methods 
suggested.                                                             §  100,  pp.  189-190. 
Higher  Roots :  figxxres  ior  X*,  x^ ;  other  methods.     Exercises  IV. 

§  101,  pp.  191-192. 

Part  II.     Simple  Powers  and  Roots  of  Polynomials. 

p.  192. 
1 
Fractional  Notation  for  Roots :  equivalence  of  x"  to  Vx ;  positive 
answer  intended  if  n  is  even.  §  102,  p.  192. 

yfonomials:  powers  —  direct  extension  of  multiplication;  roots  — 
direct  reversing  of  multiplication.     Exercises  V. 

§  103,  pp.  192-193. 


202  SIMPLE   POWERS   AND   ROOTS 

Formal  Rules ;  Powers  of  Monomials :  essentially,  raise  coefficient 
to  power,  multiply  exponents.  §  104,  pp.  193-194. 

Roots  by  Fractional  Notation  :  correctness  of  rules  of  §  104  for  exact 
roots  in  fractional  notation.    Exercises  VI.    §  105,  pp.  194-196. 

Longer  Expressions  :  powers  by  previous  rules  ;  roots  by  inspection. 
Exercises  VII.  §  106,  pp.  196-197. 

Formal  Process;  Square  Roots  of  Polynomials :  illustrative  prob- 
lems ;  key  is  "  trial  divisor  "  ;  restricted  usefulness.  Exercises 
VIII.  §  107,  pp.  197-199. 

Review  Exercises  for  Chapter  VII :  Exercises  VIII.  p.  200. 


CHAPTER   VIII.     QUADRATIC   EQUATIONS 

PART   I.     METHODS   OF    SOLUTION;   CHARACTER 
OF   THE   ROOTS 

108.  Quadratic  Equations.  If  an  equation  when  cleared 
of  fractions  and  radicals,  and  simplified,  contains  the 
square,  but  no  hio-her  power,  of  the  unknown  quantity, 
it  is  called  a  quadratic  equation,  or  an  equation  of  the 
second  degree. 

We  have  solved  some  such  equations  (see  §§  64-67, 
])p.  103-115,  and  §  82,  p.  151);  we  shall  now  show  how 
I' I  solve  any  such  equation. 

Ex.  1.    Given  the  equation  2  or —  9  x-{-4:  =  0. 

We  notice  the  factors  (2  x  -  1)  and  (x  -  -i)  on  the  left ;  hence  we 
write  (2x-- l)(j -4)  =0. 

Whence,  2  a;  —  1  =  0,  or  x  —  4  =  0, 

and  X  =  h  or  x  —  4. 

If  these  factors  were  not  noticed,  or  if  the  example  were 
so  difficult  that  the  factors  could  not  be  seen  by  inspec- 
tion, we  could  proceed  as  follows: 

Trying  various  numbers  for  x,  we  should  naturally  try  the  numbers 
a;  =  0,  X  =  1,  a;  =  2,  etc.,  to  see  if  we  could  find  a  correct  answer  by 
trial.  Letting  x  =  1,  for  example,  the  left  side  is  not  zero;  it  is 
2-9  +  4=- 3. 

Trying  various  numbers,  as  suggested,  we  should  find  a  table  like 
this  : 


X 

0 

1 

2 

3 

4 

5 

6 
2-2 

7 

etc. 

-  1 

-2 

-3 

-  4 

etc. 

I 

4 

-  3     -  6 

-  5 

0 

9 

etc. 

15 

etc. 

203 


204 


QUADRATIC    EQUATIONS 


[Cn.  VI II 


where  /  means  the  vahie  of  the  left  side  of  the  given  equation,  tliat  is, 

[Let  the  student  fill  in  the  blank  spaces.] 

These  values  may  be  plotted  in  a 
figure,  as  on  p.  183.  The  figure  shows 
the  value  of  the  left  side  for  any  value 
of  X ;  we  wish  to  have  the  left  side,  /, 
equal  to  zero,  in  order  to  satisfy  the 
given  equation. 

One  value  of  x  for  which  I  is  zero 
was  discovered  hy  trial  on  p.  20o, 
namely,  x  =  4.  It  is  clear  from  the 
figure  that  there  is  another  value  some- 
where between  0  and  1,  for  the  curve 
crosses  the  horizontal  line,  i.e.  the  left 
side  is  zero,  somewhere  between  0  and 
1,  about  .5. 

These  answers  are  correct,  as  we 
found  by  a  different  method. 

This  process  will  not  always  give  an 
accurate  answer,  of  course,  because  the 
figure  is  not  entirely  accurate. 


^ 

/ 

T 

T 

f 

F 

r 

[ 

ITa 

+  -H>       H 

1          J 

i 

^ 

•J 

j:      " 

t 

n   I- 

^l      -.,- 

__      -.i^—.J- 

^ 

\      4 

-^    t 

^    t 

.,     ,              , 

Jc  - 

\J 

X 


Fig.  35. 


EXERCISES   I  :    CHAPTER   VIII 

[The  first  of  these  exercises  are  easy  examples  by  the  factoring 
method ;  the  student  should  solve  them  by  factoring  as  on  pp.  107 
and  151.  Then  he  should  draw  the  figure  as  illustrated  above  and 
compare  the  results.]  .  j 

6.    3  ^"-7  ^-6  =  0. 


1.  a;2-7a;  +  6  =  0. 

2.  r--r-12  =  0. 

3.  «-  =  3^  +  10. 

4.  2-  =  9z-20. 

5.  2^2-11  <+!2  =  0. 


It 

7.  13j^  =  2ir  +  15. 

8.  2  7/r4-m  =  21. 

9.  10<-  +  «-3  =  0. 
10.    3  V-  =  34  V  —  40. 


[The  second  half  of  these  exercises  includes  exainples  that  the 
student  will  find  too  difficult  by  the  factoring  method.  In  these  lie 
should  only  draw  the  figure  and  estimate  the  answers.] 


Ii),s_l00]  METHODS   OF    SOLUTION  205 

11.  n- -4:11  +  2  =  0.  15.  2xr-3x-6  =  0. 

12.  771^  — 6m +  6  =  0.  16.  32-  — 7  2  +  3  =  0. 

13.  f'-5t  +  3  =  0.  17.  2j-2  +  3r-l=0. 

14.  ar  —  7  .r  —  4  =  0.  18.  x-  —  x  —  l  =  0. 

109.  General  Solution.  If  an  absolutely  precise  answer 
is  needed,  it  can  be  found  by  the  process  illustrated  below, 
no  matter  how  difficult  the  example  may  be  by  factoring. 

Ex.  1.   x'  —  Ax  —  o  =  0. 
Transpose  5  and  write,  x^  —  4:x  —  5. 

Add  4  to  each  side  :  x-  —  4  x  +  4  =  9. 

The  purpose  of  this  is  to  make  the  expression  on  the  left  side  a 
perfect  square.     Evidently  it  is  (x  —  2)'^,  and  we  have  : 

(x--J)^  =  9. 
Whence,  x  —  "2  =  ±3, 

and  X  =  2  ±  o  =  5  or  —  1. 

Check :    For  x  =  5, 

5-  —  4  •  5  —  5  =  0  (correct). 
For  X  =  —  1, 
(-  1)-  -  4  (-  1)  -  5  =  1  +  4  -  5  =  0  (correct). 

Problems  should  always  be  carefully  checked.  An 
'diswer  is  a  value  of  x  (or  whatever  letter  is  used)  that 
satisfies  the  original  equation  when  x  is  replaced  by  it. 
Such  an  answer  is  often  called  a  solution,  or  a  root  of  the 
equation. 

Ex.  2.    2 .1--  -  9  .r  +  4  =  0.  (See  Ex.  1,  p.  203.) 

Divide  by  2  on  botli  sides,  and  transpose  4 : 
x2  -  I  X  =  -  2. 

Desiring  to  make  a  perfect  square  on  the  left  side  we  arid  to  each 
side  the  square  of  \  of  the  coefficient  of  x,  that  is  (^  X  —  |)'^  =  f^, 

a^'  -  I  ^  +  f  i  =  -  2  +  H. 
or,  (X  -  1)2  =  4|. 

Whence,  ^  -  I  =  ±  1, 

or,  X  =  ?  ±  i  —  4  or  1^. 


206  QUADRATIC   EQUATIONS  [Ch.  VIII 

Check :   For  x  =  4  ;  2  •  42  -  9  .  4  +  4  =  0  (correct). 

For  x  =  i  ;  2  •  (i)2  -  9  •  (i)  +  4  =  0  (correct). 
These  are  the  same  answers  as  those  found  above  for  this  problem. 

110.    Completing  a  Square.     Remembering  the  rule 
(x  +  ay-  =  x^  +  '2  ax  -f  a^^ 

we  notice  that  the  last  term  ((fi~)  in  a  perfect  square  is  the 
square  of  ^  the  coefficient  of  x,  if  the  first  term  is  x"^.  The 
preceding  work  depends  on  this,  and  we  may  definitely 
state  the  rule  : 

To  solve  a  quadratic  equation : 

(1)  Notice  what  letter  deflates  the  unknown  quantity  ;  ive 
shall  here  call  it  x. 

(2)  Trayispose  the  terms  in  x^  and  in  x  to  the  left  side  of 
the  equation,  the  other  term  or  terms  to  the  right. 

(3)  Divide  both  sides  by  the  coefficie7it  of  x^. 

(4)  Add  to  each  side  the  square  of  half  the  coefficient  of 
X,  so  that  the  left  side  is  a  perfect  square. 

(5)  Extract  the  square  root  of  both  sides,  taking  care  to 
put  both  signs  ±  on  the  right. 

(6)  Solve  the  resulting  equations  by  transposing  to  one 
side  all  but  the  term  in  x. 

(7)  Check  each  ansiver  by  substitution  in  the  original 
equation. 

In  step  (5)  the  sign  ±  might  be  put  on  both  sides,  since  a  square 
root  is  extracted  on  each  side.  But  the  effect  is  the  same  as  that  given 
in  (.5),  since  any  equation       ±  A  =  ±  B 

with  ±  on  both  sides  is  really  the  same  as  the  equation 

A  =  ±B 
with  ±  on  one  side  only. 

[The  student  should  now  carefully  examine  the  solutions  of  the 
problems  solved  in  §  109  and  notice  that  this  rule  is  followed  there.] 

The  solution  may  not  come  out  in  rational  form,  as 
illustrated  by  the  following  example  : 


109-110] 


METHODS   OF   SOLUTION 


207 


Ex.  1.    x'-i  x  +  \=0. 

Transpose  +  1  and  add  (^  •  4)^  or  4  to  each  side: 

a;2  -  4  X  +  4  =  3. 
Take  the  sqnare  root  of  each  side : 

X  -  2  =  ±  V3, 
or,  a:  =  2  ±  V3  =  2  +  V8,  or,  2  -  V3. 

Check-:    For  2+  ^/3  : 
(2  +  V3)2  -  4  (2  +  V3)  +  1  =  4  +  4V3  +  {Vsy  -  8  -  4 V3  +  1=0. 

For  2  —  v'3  :     (Correct.) 
(2  -  Vsy  -  4  (2  -  v'3)  +  1  =  0.    (Correct ;  student  finish  the  work.) 

The  figure  is  drawn  as  in  §  108. 


r    0 

1 

2 

3 

4 

5 

6 

7 

8 

etc. 

-  1 

_  2 

-3 

-4 

etc. 

/    1 

_  o 

-  3 

_  2 

1 

6 

6 

]'"rom  the  figure  the  answers  are  seen  to  be 
X  =  0.3  (about), 
and  X  =  3.7  (about). 

This  compares  well  with  the  results  above, 
for  Vij  =  1.73+ ;  hence  the  answers  found 
above  are 

X  =  2  -  1.73+  and  x  =  2  +  1.73+, 
or,      X  =  .27+  and  x  =  3.73+. 

It  is  Keen  that  the  figure  shows  approxi- 
mately these  square  root  values. 

EXERCISES   II:    CHAPTER   VIII 

[In  solving  these  exercises,  the  student 
should  draw  a  figure  for  each  one  as  in 
§  108.  The  answers  found  from  the  figure 
will  serve  as  a  check  on  the  correctness  of 
the  work  of  solution.] 

1.  0.-2-15.^  +  44  =  0.  4.    x" 

2.  jj2  _  15  ^,  4,  54  ^  0.  5 

3.  a^- 15  a  =  100. 


^; 


X!^.A-X 


Fig.  36. 
15  X  =  76. 

ff  -  15  d  =  154. 
6.    2 r^-r  =  36. 


208  QUADRATIC   EQUATIONS  [Ch.  VllI 

7.  10n2_3i^i  +  24  =  0.  17.    m2  +  6m  =  3. 

8.  3  a- -11  a +  10  =  0.  18.  ^^-10^  =  5. 

9.  4a;2  +  4.«=15.  19.   x^  +  x  =  7. 

10.  x^  —  7x  +  3  =  0.  20.  x'2  +  3a;  =  10. 

11.  .T--5.T  +  3  =  0.  21.  2ic2  +  5x  =  4. 

12.  9<'-3n-3  =  0.  22.  3j>2_^4j>  =  6. 

13.  x^-x-3  =  0.  23.  a2x2  +  2ate  =  a2-62. 

14.  5"  +  <7  —  3  =  0.  (Do  not  plot  figure.) 

15.  r-  +  3j--3  =  0.  24.  m2  +  llm  =  3. 

16.  a-  +  2a  =  14.  25.  6a-2  +  3aj  =  10. 

Also  solve  the  exercises  in  List  I,  pp.  204-205,  by  the  method 
of  §  110  and  compare  with  the  answers  found  before. 

111.  Second  Method.  Another  method  slightly  different 
from  the  preceding  will  now  be  illustrated  by  the  ex- 
ample used  in  §  109  (Ex.  2). 

Ex.  1.    2a.-2-9a;  +  4  =  0. 

Multiply  each  side  by  8,  and  transpose  32 : 
16  x2  -  72  X  =  -  32, 
or,  (4j)2_  18(4  a.)  =  -32. 

Add  the  square  of  J  the  coefficient  of  4  x  in  the  middle  term,  that 
is,  the  square  of  i  of  18,  or  9'-^ : 

(4  xy  -  18(4  x)  +  (9)2  =  _  32  +  (9)2, 
or,  (4x  -  9)2  =  49; 

therefore,  4a:  =  9±7  =  16or2, 

whence,  x  =  4  or  |. 

These  are  the  same  answers  as  were  found  before. 

This  method  is  practically  the  same  as  the  previous  one,  except 
that  4  X  is  used  instead  of  x  in  performing  the  operations. 

This  method  is  sometimes  more  convenient,  since  long 
fractions  may  be  avoided.  It  will  be  enougli  to  start  hy 
muUipIi/inff  both  sides  hy  four  times  the  coefficient  of  x^,  so 
that  the  term  in  jfi  is  an  easy  perfect  square. 


1] 0-112]  METHODS   OF    SOLUTION  209 

This  method  is  often  called  the  Hindu  method,  owing  to  its  origin 
in  India.  In  many  cases  the  work  may  be  shortened  by  multiplying 
l)y  a  smaller  number  than  four  times  the  first  coefficient.  Thus,  the 
solution  of  the  preceding  example  may  be  shortened  somewhat  by 
multiplying  by  2  instead  of  by  8  and  then  solving  in  terms  of  (2  x) 
in  place  of  (Ix).     Let  the  student  do  this. 

It  is  best  to  use  the  method  previously  given  as  the 
standard  method,  and  to  use  this  new  one  only  rarely. 

EXERCISES   III:   CHAPTER  VIH 

Solve  the  following  examples  both  by  the  method  of  §  110 
arid  by  the  method  of  §  111,  comparing  the  two  solutions  for 
ease  and  accuracy.     Abbreviate  the  latter  method  when  pos- 
sible.    Always  draw  a  figure  as  a  check. 
'     1.    2a;2-llx  +  12  =  0.  10.    5.T--a;-18  =  0. 

2.  2^/-3i>-5  =  0.  11.   3 <-- 8^-7  =  0. 

3.  2z^-nz-9  =  0.  12.   4:Z--3z-3  =  0. 

4.  ;w--ll<  +  G  =  0.  13.   2r-/--4  =  0. 

5.  :u'-llt-4:  =  0.  14.   3r-2r-3  =  0. 

6.  3  q--{-q  — 10=0.  15.    5?»- —  3  7?i  —  4  =  0. 

7.  4H2-ir>n  +  15  =  0.  16.    9.^'--10.^•-3  =  0. 

8.  4jr-23n  +  lo  =  0.  17.    7x-- -  12  .x-  + 3  =  0. 

9.  6L--L-15  =  0.  18.   3z2_7z  +  3  =  0. 

^/112.  Equal  Roots.  Instead  of  having  two  difFerent 
answers,  as  in  most  of  the  examples  above,  a  quadratic 
equation  may  have  only  one  ansiver. 

Ex.  1.    x^  _  4  X-  +  4  =  0. 

The  left  side  is  already  a  perfect  square ;  taking  the  square  root 

of  each  side : 

x  -  2  =  0,  or   x  =  2, 

since   Vo  is  0. 

This  equation  has  only  one  solution.     It   is  clear  that 
this  will  be  true  whenever  all  the   termn  form  a  perfect 
square,  after  they  have  been  transposed  to  one  side. 
hedkick's  el.  alg.  —  14 


210 


QUADRATIC   EQUATIONS 


[Ch.  VIII 


It  is  sometimes  said,  in  this  case,  that  the  equation  has 
two  equal  roots,  in  order  to  keep  up  the 
notion  of  two  roots. 

The  figure  in  this  case  has  only  one 
point  on  ihi  liorizontal  line,  for  there 
is  only  one  value  of  x  which  makes  the 
left  side  =  0. 

The  corresponding  values  of  x  and  of 
/  =  x"^  —  4  X  +  4  are : 


1. — 

_   _   _.. 

- 

S- 

-r- 

__ 

-----\---l- 

==E=:== 

— 4--!^- 

1--- 

_ 

:::=h:: 

:/::::: 

r^ 

:::=:\:: 

/:::::: 

t:::o±d 

— a 

IJT- 

^4«-;-4- 

^ 

X 

I 

0 
4 

1 

1 

2 

0 

3 

1 

4 

4 

5 

etc. 

-  1 

etc. 

etc. 

9 

etc. 

Fig.  37. 


Plotting  these  values  in  a  figure  (Fig.  37), 
as  in  §  108,  we  find  that  only  one  point,  tlie 
point  for  which  x  =  2,  is  on  the  horizontal 
line.  Hence  there  is  but  one  answer,  x  =  3, 
as  found  above. 


iyll3.   Imaginary  Roots.     It  may 

also  happen  that  a  quadratic  equa- 
tion has  no  solution. 

Ex.  1.    a^  +  2  =  0. 

Drawing  the  figure,  as  before,  we  find : 


X 

I 

0    12 
2    3     6 

etc. 

-  1 

_  2 

-  3 

etc. 
etc. 

3 

6 

11 

From  the  figure,  and  from  the  work 
done  in  making  the  table,  it  is  clear  that 
there  is  no  caluc  of  x  among  the  numbers 
familiar  to  us  for  which  the  left  side  is 
zero ;  for  the  smallest  value  of  Ms  2  (for 
X  =  0),  and  the  value  of  /  rises  steadily 
from  this  lowest  value  on  both  sides. 


zi5 


-■i 


=  Xijt-  ^ 


X 


Fig.  38. 


112-113] 


METHODS   OF   SOLUTION 


211 


Ex.  2.    x2_^2.t4-5  =  0. 
Drawing  as  before,  we  find 


./■  1  0    1 

/'77 

2 

13 

3 

4 

etc. 

-1 

_  2 

etc. 

20 

4 

5 

etc. 

--5 


I 


Fig.  39. 


Here,  again,  the  value  of  I  is  never 
zero  ;  it  is  smallest  for  x  =  —  1  and  rises 
steadily  on  each  side  of  x  =  —  1. 

These  examples  show  that  it  is 
possible  for  a  quadratic  equation 
to  have  no  solution  among  num- 
bers we  know.  Care  must  be 
taken  to  draw,  the  figure  very  ac- 
curately, for  a  slight  error  might 
make,  it  appear  that  the  figure  for 
example  1,  §  112,  did  not  have  a  point  on  the  horizontal  line. 

Referring  to  Ex.  1,  the  numerical  work  may  be  done 
as  follows  : 

Ex.  1.   .^-'  +  2  =  0. 

Transpose  "2  :  x^  =  —  2  or  x  =  V—  2. 

But  there  is  no  ordinary  number  whose  square  is  equal 
to  —  2,  for  the  squares  of  the  numbers  we  know  are  all 
positive.     Hence,  there  is  no  solution. 

Ex.2.    a^  +  2a;  +  5==0. 

Proceeding  as  if  we  were  solving  by  the  usual  method,  we  transpose 
5  and  add  the  square  of  |  •  2  to  each  side  : 

x2  +  2  ar  +  1  =  -  4,        

or,  /x  +  1)2  =  -  4,  or  X  =  -  1  ±  y/^^. 

But  this  answer  is  meaningless,  for  we  know  no  number  whose  square 
is  —  4.  In  such  an  example  the  expressed  answers  (in  this  example 
_1  ^--v/ZTl)  are  meaningless  to  the  student  at  present;  they  are 
often  called  imaginary  solutions.  Later  on  (see  Appendix,  §  31),  a 
certain  useful  meaning  will  be  given  to  them. 

There  follow  several  examples,  some  of  which  have  just 
one  solution,  and  some,  no  solution. 


1. 

2r- 5}) +  8=0. 

2. 

x'-4.x  +  5  =  0. 

3. 

a^_3aj  +  4  =  0. 

4. 

^2-6^  +  9  =  0. 

5. 

2^+ 72 +  13  =  0. 

6. 

m-  -  9  m  +  21  =  0 

7. 

x2-8x  +  16  =  0. 

8. 

^2-3^  +  3  =  0. 

9. 

22-42  +  8  =  0. 

212  QUADRATIC   EQUATIONS  [Ch.  Mil 

EXERCISES   IV:   CHAPTER  VIII 

[Draw  the  figure  in  each  case ;  also  make  an  attempt  at  solu- 
tion, as  above.] 

11.  2L'-7L  +  7  =  0. 

12.  3m2-10m  +  9  =  0. 

13.  4?r-12n  +  9  =  0. 

14.  5/)-  — 6;? +  3  =  0. 

15.  2a^-4.T  +  3  =  0. 

16.  9s2-30s  +  25  =  0. 

17.  e-12t  +  S6  =  0. 

18.  .T^- 7.^  +  13  =  0. 

19.  2/-5?/  +  4  =  0. 
10.    >-2-8r  +  20  =  0.                    20.  3m2-4m  +  3  =  0. 

114.  General  Solution  by  Formula.  We  shall  now  ex- 
press the  solution  of  any  quadratic  equation.  The  stu- 
dent has  seen  above  that  we  could  always  do  so,  though 
the  expressed  solution  may  not  have  a  meaning. 

Any  quadratic  equation  is  of  the  form : 

ax^  +  bx  +  c  =  0, 

where  a,  b,  c  are  fixed,  known  numbers  in  any  one  example, 
and  a  is  not  zero.     Divide  both  sides  by  «,  as  on  p.  206. 

a^  +  ^x+'^-=0. 
a        a 

c  f  h\^ 

Transpose  -  and  add    —  )  to  both  sides  : 

a  V2  aj 

a        \z  a)        \2  a)       a 
The  left  side  is  found  to  be  the  square  of  (a;  -I ) : 

"^      -laj  4  a2 


l.j-114]  METHODS   OF    SOl.LTION  213 

Taking  the  square  root  of  each  side,  we  find 


:r+A=±J*!l4^=±:^^^!^^   (See  §99,  p.  188.) 


Transposing-  -—,  we  find 


_       b       VA-  —  4  ac 
2a  2a 


-b±  y/b'-^ac 

^  = 2  a • 

Check:  These  answers  are  correct,  for  we  may  reverse  each  step, 
starting  with  the  answers  and  going  backwards  through  the  work  to 
tiie  original  equation.  Let  the  teacher  guide  the  students  in  doing 
this,  accounting  for  each  step  as  it  is  taken. 

The  answers  may  also  be  checked  by  direct  substitution  in  the 
given  equation,  ax'^  +  bx  +  c  =  0.  This  work  is  not  written  here  be- 
cause it  is  long.  The  teacher  should  guide  the  students  in  actually 
doing  it. 

Ex.  1.   x'-4x-5  =  0.     (See  Example  1,  §  109.) 

Comparing  with  ax'^  +  bx  +  c  =  0,  we  find 

a  =  1,  b  =  —  i,  c  =  —  5. 
Putting  these  numbers  in  the  general  result  just  found,  we  get 


(_4)±V(-4)^-4(l)(-5) 
2(1) 

2  2 

which  are  the  answers  found  before. 

Ex.  2.   2x'-9x  +  A  =  0.     (See  example  2,  §  109.) 

«  =  2,  i  =  -  9,  c  =  4.     .-.  b-'  -  4  ac  =  (-  9)2  -  4  (2)(4)  =  49, 

,^IlXil9)±V49^9^^4^ 
2-2  4 

It  is  often  best,  as  here,  first  to  calculate  the  value  of 
52  —  4  ac,  the  quantity  under  the  radical  sign.  One  reason 
for  this  is  seen  in  the  next  article. 


214 


QUADRATIC    EQUATIONS 


[Cir.  VIII 


The  other  examples  solved  in  the  text  above  are  given 
in  the  following  table,  solved  by  the  preceding  formula. 


\u 

KXAMPLK 

(( 

b 

c 

b^-iac 

J-  (not  reduced) 

X'  (reduced; 

3 

a;2  -  4  X  +  1  =  0 

1 

-4 

1 

12 

(4  ±  \/l2)  -  2 

■  2  ±  v'3 

4 
5 

x2  _  4  jc  +  4  =  0 

1 

-4 
0 

4 

2 

5 

0 

(4±V0)-2 

2  (no  other) 

x2  +  2  =  0 

1 

-8 

(0±V^H-2 

meaningless 

6 

x2  +  2x  +  5  =  0 

1 

2 

-  16 

(2±\/-  16)^2 

meaningless 

115.  Discriminant.  After  a  little  practice  it  will  be 
noticed  that  the  value  of  6^  —  4  ac  shows  the  nature  of  the 
result. 

A.  1.  If  i^  —  4  ac  >  0,  there  are  two  answers  ("  unequal 
real  roots  "). 

(In  a  figure,  drawn  as  in  §  108,  the  curve  cuts  the  main  horizontal 
in  two  points.) 

A.  2.  If  Z>2  —  4  ac  =  0,  there  is  only  one  answer  ("  equal 
roots  "). 

(In  a  figure,  drawn  as  in  §  112,  the  curve  just  touches  the  main 
horizontal  line.) 

A.  3.    If  6^  —  4  ac  <  0  there  are  no  answers  ("  imaginary 

roots'''). 

(In  a  figure,  drawn  as  in  §  113,  the  curve  does  not  cut  the  main 
hoiizontal.) 

For  b^  —  iac  is  the  quantity  under  the  radical  sign  ;  if  it  is  negative 
we  have  the  square  root  of  a  negative  quantity,  which  is  meaningless 
to  the  student  at  present. 

B.  If  b^  —  4  ac  is  an  exact  perfect  square,  the  answers  are 
rational;  otherwise,  the  ansivers  are  irrational,  provided  a, 
h,  e  are  rational. 


Ill   n.->]  METHODS   OF    SOU'TfOX  215 

Oil  account  of  these  facts  the  quantity  //-  -  -i  uc  is  often  called  the 
discriminant;  the  knowledge  of  its  value  enables  us  to  discriminate 
among  the  cases  mentioned. 

Notice  that  if  these  rules  are  forgotten,  all  this  information  can  be 
tiMiiid  in  any  one  example  by  attempting  to  solve  as  in  §§  109-113, 
l)y  completing  the  square. 

EXERCISES    V:    CHAPTER  VIII 

Draw  the  figure,  solve  by  completing  the  square,  and  also 
solve  by  the  formula.     Compare  the  three  answers. 

1.  X-  -  10  X  +  16  =  0.  9.  0  7-  -f  /■  -  2  =  0. 

2.  .3.t--  +  5.K-12  =  0.  10.  5  2- -3  2 -36  =  0. 

3.  2  2- -9:2 +  4  =  0.  11.  3jr-14H-5  =  0. 

4.  :iy--8y  +  rj  =  0.  12.  O2r-17p  +  G  =  0. 

5.  2t--\-  t-  15  =  0.  13.  7  x'^  -  10  a-  +  3  =  0. 

6.  6ii2_i3w  +  6  =  0.  14.  9  7.- -30  7. +  25  =  0. 

7.  '3o--22v  +  7  =  0.  15.  8  fc- -  2  A; - 15  =  0. 

8.  4  »r  -  20  m  +  25  =  0.  16.  8  x-  -  x  -  30  =  0. 

Pind  by  calculating  &-  — 4ac  whether  the  roots  are  (A)  real 
and  unequal,  equal,  imaginary  ;  (B)  rational  or  irrational,  with- 
out actually  finding  the  roots. 

17.  5  7---7r-6  =  0.  27.  6--ll.s  +  32  =  0. 

18.  4. S-- 28  s +  49  =  0.  28.  <-  +  8^  +  15  =  0. 

19.  2  7/r  +  /u-21=0.  29.  4^--12'<;  +  9  =  0. 

20.  3p-  +  4^>-6  =  0.  30.  6.^-  +  13.^;  +  8  =  0. 

21.  2.(r-lla;  +  12  =  0.  31.  3  ?/--?/ -1=0. 

22.  3  ir  —  7  />  +  5  =  0.  32.  4  r-  —  7  r  +  5  =  0. 

23.  3  r  -  7  /•  +  4  =  0.  33.  3  I?  -  L  +  1  =  0. 

24.  s-  +  9  .s  +  1 0  =  0.  34.  X-  —  {m  +  1 )  .r  +  //(  =  0 . 

25.  4  ?/-  —  11  ?/  +  9  =  0.  35.  as-  +  (a  +  h)  .s  +  ?>  =  0. 

26.  16?/r-8m  +  l  =  0.  36.  4  a^X"  +  4  a.r  +  1  =  0. 


PART    II.     PRACTICAL   APPLICATIONS;    PROBLEMS 

116.  Practical  examples  have  been  given  that  lead  to 
quadratic  equations  (see  pp.  108-112).  The  following 
may  now  be  solved : 

EXERCISES    VI  :     CHAPTER   VIII 

After  forming  the  equations,  solve,  and  draw  the  graph : 

1.  The  sum  of  the  squares  of  two  consecutive  integers  is  61. 
What  are  the  numbers  ? 

2.  A  number  is  less  by  6  than  the  third  of  its  square. 
What  is  the  number  ? 

3.  Find  two  consecutive  integers  whose  product  is  462. 

4.  What  numbers  differing  by  7  have  the  product  60  ? 

5.  Find  a  number  of  two  digits,  whose  tens'  digit  exceeds 
its  units'  digit  by  one,  if  the  sum  of  the  digits  times  the 
original  number  is  63. 

6.  Find  a  number  that  exceeds  twice  its  square  by  ^. 

7.  Find  three  consecutive  integers  such  that  the  sum  of  the 
squares  of  the  first  two  is  the  square  of  the  third. 

8.  Find  a  number  whose  square  multipled  by  3  exceeds  20 
times  ,the  number  by  7. 

9.  Find  two  numbers  whose  sum  is  12,  and  the  sum  of 
whose  squares  is  74. 

10.  Find  three  consecutive  odd  integers  such  that  the  first 
two  taken  in  order  as  the  digits  of  a  number  express  the  prod- 
uct of  the  last  two.    Solve  the  same  problem  in  even  integers. 

11.  A  stream  flows  at  the  rate  of  5  miles  an  hour;  a  crew 
rows  6  miles  with  the  stream  and  the  same  distance  back  in  3^ 
hours.     What  is  the  rate  of  the  boat  in  still  water  ? 

12.  A  crew  rows  upstream  against  a  current  of  3  miles  an 
hour,  for  a  distance  of  8  miles;  then  back  again.  If  the  trip 
takes  5  hours,  wliat  rate  could  the  crew  make  in  still  water  ? 

216 


PKACriCAL    ArrLRATIUXS;    PROBLEMS  217 

13.  A  crew  rows  upstreain  4^  miles  against  a  current  of  4 
miles  an  hour ;  it  drifts  back  2  miles,  and  rows  at  the  original 
rate  to  the  starting  point.  What  is  the  rate  of  rowing,  if  the 
trip  is  made  in  5  hours  ? 

14.  A  crew  able  to  make  3  miles  an  hour  in  still  water  rows 
8  miles  upstream  and  back  in  a  total  time  of  6  hours.  What 
is  the  rate  of  the  current  ? 

15.  A  crew  able  to  make  5  miles  an  hour  in  still  water  rows 
6  miles  upstream  and  drifts  back  in  a  total  time  of  5  hours. 
How  fast  is  the  current  flowing  ? 

16.  An  open  box,  to  be  made  from  a  square  piece  of  card- 
board, as  in  example  1,  p.  103,  by  cutting  out  a  four-inch  square 
from  each  corner  and  turning  up  the  sides,  is  to  contain  256 
cubic  inches.     How  large  a  square  must  be  used  ? 

17.  How  large  a  square  should  be  cut  from  each  corner  of  a 
piece  of  tin  18  in.  square,  to  form  an  open  box  whose  total  sur- 
face area  is  260  sq.  in.,  by  turning  up  the  sides  ? 

18.  A  piece  of  cardboard  5  in.  longer  than  wide  is  used  to 
make  a  box  of  capacity  108  cu.  in.  by  the  method  of  Exs.  16, 17. 
How  large  a  card  must  be  used  if  3-inch  squares  are  cut  from 
the  corners  ? 

19.  From  a  card  three  times  as  long  as  wide  a  box  containing 
56  cu.  in.  is  made  as  above  by  cutting  half-inch  squares  from 
the  corners.     Find  the  size  of  the  card. 

20.  From  what  size  of  card,  two  thirds  as  wide  as  it  is  long, 
can  a  box  containing  125  cubic  inches  be  made  by  cutting  out 
squares  of  side  21  inches  from  the  corners  ? 

21.  If  the  error,  in  inches,  made  in  measuring  the  side  of 
a  scpiare  10  ft.  long  is  denoted  by  e,  and  the  resulting  error, 
in  square  inches,  in  the  computed  area  by  E,  E  =  e~-^24:0  e. 
(See  example  2,  p.  108.)  Find  ^iff^  =  l;  3;  i;  0;  -1;  etc. 
Plot  the  graph.  Find  e  if  ^  =  484 ;  if  £  =  1000 ;  if  ^  =  -  711. 
Solve  for  e  in  terms  of  E,  in  general. 


218  QUADRATIC    EQUATIONS  [Ch.  VIII 

22.  In  Ex.  21,  find  E  in  terms  of  e  if  the  side  of  the  square 
is  25  ft.  long.  Plot  a  figure.  Find  e  if  ^  =  2500.  Solve  for 
e  in  terms  of  E,  in  general. 

23.  In  Ex.  22,  how  nearly  must  the  side  be  measured  to 
make  the  computed  area  correct  to  within  2  sq.  ft.  ?  If  a  foot 
rule  is  used,  how  accurately  must  the  ruler  be  placed  each 
time  ?     Is  this  practicable  ?     See  Exs.  51-53,  p.  115. 

24.  If,  in  Ex.  21,  the  square  is  replaced  by  a  circle  whose 
radius  is  10  ft.,  answer  the  same  questions  asked  in  Ex.  21, 
assuming  that  the  radius  is  measured.     See  Ex.  18,  p.  112. 

Boyle's  Law  states  that  under  constant  temperature  the  product  of 
the  volume,  v,  and  pressure,  p,  of  a  quantity  of  gas  remains  constant. 
This  is  true  whatever  units  of  measurement  are  used ;  we  use  "  centi- 
meters of  mercury  "  for  pressure  and  liters  for  volume.  The  product 
"/)y  "  in  those  units  will  be  called  A'. 

25.  A  mass  of  hydrogen  for  w^hich  /t'=  85,120  is  confined 
in  a  collapsible  reservoir  of  unknown  capacity.  An  increase 
of  the  pressure  by  4  centimeters  causes  the  reservoir  to  collapse 
partially,  decreasing  the  volume  by  56  liters.  What  was  the 
original  pressure  ?     The  original  volume  ? 

Solution.     Let  p  be  the  original   pressure.     Then    - — ^^-  is  the 

P 
original    volume.      The   new   pressure    is   p  +  4:,   the    new   volume 

8.5120      ^f,      XT 
56.     Hence, 


(p  +  4)  (55120  _  5g\  ^  ggj2o, 


P 

p2+  ij)  -6080  =  0, 
or,  on  solving,  79  =  —  80  or  76. 

The  solution  of  the  problem  is  ^  =  76  centimeters ;  also,  we  see 
easily  that  v  —  1120  /. 

26.  For  a  certain  mass  of  gas  in  a  collapsible  reservoir, 
K  =  277,500.  A  decrease  of  1  cm.  in  ]iressure  causes  an  increase 
of  50  /.  in  volume.     Find  the  original  volume  and  pressure. 


IHJ]  PRACTICAL   APPLICATIONS;     PROBLEMS  219 

Solve  the  following  similar  problems: 

27.  K  =  162,0(50.     Increase  in  p,  1  cm.     Decrease  in  v,  30  Z- 

28.  K  =  275,025.    Increase  in  }>,  1-5  cm.    Decrease  in  v,  75  /. 

29.  /r=  112,480.     Decrease  in  p,  2  cm.     Increase  in  v,  40  ^. 

30.  Two  tanks  of  equal  capacity  are  emptied  by  unequal 
pipes;  it  is  observed  that  the  tank  having  the  larger  pipe  is 
empty  two  hours  sooner  than  the  other.  JJoth  pipes  attached 
to  a  single  tank  till  it  in  24  hours.  How  long  do  the  pipes 
separately  require  to  fill  or  empty  the  tanks? 

31.  Two  equal  tanks  are  filled  by  pipes,  one  requiring  3  hours 
longer  than  the  other.  Both  pipes  together  fill  one  tank  in  2 
hours.     How  long  does  each  pipe  require  to  fill  one  tank  ? 

32.  Two  pipes  are  used,  the  larger  to  fill  a  tank,  the  smaller 
to  empty  it.  When  they  are  both  open,  the  tank  is  filled  in 
1 5  hours.  The  pipes  are  then  detached  and  used  separately  to 
empty  two  tanks  each  equal  to  the  first.  It  is  observed  that 
this  work  is  done  4  hours  sooner  by  the  large  than  by  the 
small  pipe.     Find  the  number  of  hours  required  by  each  pipe. 

33.  A  tank  is  filled  by  a  certain  pipe  in  an  hour  less  than 
is  required  for  a  larger  pipe  to  fill  a  tank  of  twice  the  capacity 
of  the  first.  Both  pipes  together  fill  the  large  tank  in  3  hr. 
45  min.     In  what  time  could  each  pipe  fill  the  small  tank  ? 

34.  Two  tanks  whose  capacities  are  as  2  to  3  are  emptied 
by  pipes,  the  larger  tank  requiring  an  hour  longer  than  the 
other.  Both  pipes  together  fill  the  larger  tank  in  1^  hr. 
How  long  would  each  separately  require  to  fill  the  small  tank  ? 

35.  Two  men  are  on  streets  at  right  angles  to  each  other, 
distant  7  and  8  feet  from  the  crossing.  If  they  approach  the 
corner  at  the  same  rate,  how  far  must  each  walk  so  that  their 
distance  apart  shall  be  5  feet  ?     Comment  on  the  two  answers. 

36.  Two  men  standing  1  foot  and  8  feet  from  the  crossing 
of  two  streets  at  right  angles  to  each  other  walk  away  from 
the  crossing  at  the  same  rate.  When  will  they  be  13  feet 
apart  ?     Interpret  the  two  answers. 


220 


QUADRATIC   EQUATIONS 


[Cn.  vin 


37.  Two  men  stand  on  streets  meeting  at  right  angles,  in 
positions  3  and  5  feet  from  the  crossing.  How  far  must  each 
walk  toward  the  crossing  at  the  same  rate  in  order  that  they 
may  be  10  feet  apart?  Interpret  the  two  answers.  Here 
neither  really  satisfies  the  implied  conditions  of  the  problem. 

38.  Two  electric  cars  are  on  tracks  meeting  at  right  angles. 
One  starts  from  the  crossing  at  the  rate  of  20  feet  a  second, 
the  other  starts  100  feet  away  from  the  crossing  at  the  rate  of 
10  feet  a  second.     When  will  they  be  200  feet  apart  ? 

A  concave  mirror  is  formed  from  a  part  of   a  spherical  shell  of 

radius  r.  If  the  distance  of  an 
object  (0)  from  the  mirror  is 
denoted  by  u,  and  the  distance  of 
the  image  (/)  of  the  object  from 
the  mirror  is  denoted  by  v,  it  is 
known  that  approximately 

n  V  r 
(Tlie  proof  of  this  formula  need 
not  be  attempted ;  being  an  ap- 
proximation, its  derivation  is  not 
a  direct  proof.)  Note  that  image 
and  oliject  are  situated  on  oppo- 
site sides  of  the  center  (C)  of  the 
mirror,  also  that  placing  the  ob- 
ject where  the  image  was,  throws  the  image  where  the  object  was. 

39.  How  far  is  an  object  from  a  concave  mirror  of  radius 
20  cm.  if  its  reflection  is  15  cm.  farther  from  the  mirror  ? 

40.  Where  are  the  object  and  its  reflection  in  the  mirror  of 
Ex.  39,  if  their  distance  apart  is  48  cm.  ? 

41.  What  radius  must  be  chosen  for  a  concave  mirror  in 
order  that  an  object  7^  cm.  from  the  mirror  may  be  reflected 
5  cm.  farther  from  the  mirror  than  the  center  ? 

42.  An  object  is  placed  beyond  the  center  of  a  concave  mirror 
so  that  the  distance  from  the  object  to  the  center  of  the  mirror 
shall  be  twice  the  distance  from  the  center  to  the  reflection. 
Where  is  the  object  placed  ?  (State  the  result  in  terms  of  the 
radius  of  the  mirror.)     Comment  on  the  two  results. 


lid]  rilAUJlCAL    Ari'LICATIU>'6;     TUUBLEMS  221 

43.  I  desire  to  place  an  object  before  a  concave  mirror  so 
that  the  center  of  the  mirror  may  lie  halfway  between  the 
object  and  its  image.     Can  I  do  so  ? 

44.  A  cylindrical  box  6  inches  high,  open  at  the  top,  has  a 
surface  of  (54  tt  sq.  in.     What  radius  must  be  chosen  ? 

45.  If  the  box  of  Ex.  44  is  6  in.  high  and  if  the  radius  of  its 
l)ase  is  4  in.,  show  that  an  error  of  e  (in  in.)  in  measuring  the 
radius  causes  an  error  E  (in  sq.  in.)  in  the  total  surface,  such 
that  E  =  7r  {20e-\-  e-).     Find  e  if  ^  =  21  tt  ;  if  ^  =  5  tt. 

46.  About  how  much  error  e  woidd  be  caused  if  we  took 
TT  =  3  in  place  of  tt  =  3|  in  the  value  64  tt  mentioned  in  Ex.  44? 

47.  A  solid  cylinder  41  inches  high  is  entirely  covered  with 
45  TT  square  inches  of  paper.     Find  the  radius  of  the  base. 

48.  Show  that  the  error  E  in  the  computed  total  area  of 
the  cylinder  of  Ex.  47,  caused  by  an  error  e  in  measuring  the 
radius,  [sE=(2  <?-  +  21  e)w.  Find  e  if  i'  =  50  tt  ;  if  J^  =  -  45  tt. 
Solve  for  e  in  terms  of  E  in  general. 

49.  Find  the  radius  of  the  base  of  a  cone  whose  slant  height 
is  7  cm.  and  whose  total  surface  area  is  800  tt  sq.  cm. 

50.  A  box  which  is  open  at  the  top  is  to  be  constrncted  20 
inches  high,  with  a  square  base.  What  are  the  dimensions  of 
the  base,  if  the  surface  of  the  box  is  1530  square  inches  ? 

51.  If  an  error  e,  in  inches,  is  made  in  measuring  the  side 
of  the  base  in  Ex.  50,  show  that  the  error  E  in  the  computed 
area  of  the  box,  in  square  inches,  is  E  =  e-  -{- 192  e.  Find  E  if 
e  =  1 ;  2  ;  3 ;   - 1 ;   etc.     Plot  the  graph.     Find  e  if  ^  =  985. 

52.  How  nearly  must  the  measurement  side  of  the  base  in 
Ex.  51  be  made  in  order  that  the  computed  area  may  be  correct 
to  within  1  sq.  ft.  ?    Is  it  practicable  to  do  this  with  a  foot  rule  ? 

53.  A  solid  block,  10  in.  high,  is  to  be  twice  as  long  as  it  is 
wide.     Find  its  dimensions,  if  the  surface  area  is  1008  sq.  in. 

If  a  body  is  dropped  from  a  point  near  the  earth's  surface,  the 
number  of  feet  it  falls  in  t  seconds  is  given  by  the  relation  .s  =  16  A 
If  instead  of  being  merely  dropped,  the  body  is  thrown  downward  at  a 


222  QUADRATIC   EQUATIONS 

speed  of  v  feet  a  second,  the  relation  is  .s  =  16  /^  +  vt.  If  the  body  is 
thrown  upward,  the  relation  is  s  =  16  t'^  —  vt.  Notice  that  in  the  last 
instance,  since  s  represents  distance  doivnward,  and  the  body  starts 
upward,  s  will  be  negative  at  first,  until  t  =  v  ^  16. 

54.  A  hotly  is  thrown  downward  with  a  speed  of  50  feet  a 
second  from  a  distance  of  225  feet  from  the  earth.  When 
will  it  strike  the  ground  ? 

55.  When  will  a  body  thrown  upward  with  the  same  speed 
from  the  same  place  as  in  Ex.  54,  strike  the  ground  ? 

56.  When  will  a  body,  thrown  as  in  Ex.  55,  descend  to  the 
level  from  which  it  was  thrown  ?    Comment  on  the  two  answers. 

57.  When  w^ill  a  stone,  thrown  downward  at  a  speed  of  74 
feet  a  second  from  a  height  of  one  mile,  reach  the  earth  ? 

58.  If  the  error  in  measuring  the  time  of  fall  of  a  body 
which  is  dropped  from  a  height  is  e,  in  seconds,  show  that  the 
error  in  the  computed  value  of  the  distance  fallen  through  is 
E  =  o2te-\- 16  e-,  where  t  is  the  real  time  of  fall.  If  t  =  10, 
E  =  320  e  +  16  e\  Find  ^  if  e  =  1,  2,  i  |,  - 1,  -  2,  etc.  Plot 
the  graph.  Find  e  if  £"  =  516.  How  carefully  must  the  time 
be  measured  in  order  that  the  error  in  the  com})uted  distance 
may  be  less  than  50  ft.  ?  How  nearly  can  the  distance  be  com- 
puted with  a  stop  watch  which  measures  to  fifths  of  a  second  ? 

59.  A  body  is  thrown  upward  from  a  height  of  1728  feet 
at  a  speed  of  48  feet  a  second.  When  will  it  reach  the  earth  ? 
When  does  it  reach  the  level  from  which  it  is  thrown  ? 
How  much  time  ela])ses  between  its  reaching  the  level  from 
which  it  is  thrown  and  its  reaching  the  earth  ? 

With  what  speed  should  it  be  thrown  doivntvard  in  order 
to  reach  the  earth  in  this  time  ?     Comment  on  tlie  result. 

60.  Show  that  the  error  E  in  the  computed  value  of  the 
distance  fallen  through  in  Ex.  54,  caused  by  an  error  e  in 
measuring  the  time  of  fall,  is  E  =  130  e  -\- i6e-.  Plot  the 
graph.  Find  e  if  £"  =  69  ;  if  E  =  —  159.  How  nearly  can  the 
distance  be  measured  by  means  of  a  stop  watch  that  reads  to 
fifths  of  a  second? 


I 


PART    III.     PROPERTIES    OF    QUADRATIC    EQUATIONS 

117.  Given  Roots.  Factor  Theorem.  We  can  now 
manufacture  eij^uations  which  shall  have  any  roots  \\<i 
wish. 

Ex.  1.    If  2  .«r  -  9  .j;  +  4  =  0, 

(,;_4)(2.:-l)=0. 
Hence,  a;  —  4  =  0,  or,  2x  —  1  =  0, 

that  is,  X  =  i,  or,   x  =  h. 

Suppose  we  wish  to  malce  an  equation  with  roots  4  and  I,  then  we 

and  find  the  product  of  these  factors.     We  get 

x^-^x  +  2^0, 
or,  multiplying  both  sides  by  2  to  clear  of  fractions,  we  get 

•2  x-^  -  9  a;  +  4  =  0. 

In  general^the  roots  will  be  r  and  s  if  the  equation  is 
(•r-r)(.r-s)=0, 
or,  a^  —  {r  +  s^x -\- rs  =  0.  (See  §  60.) 

To  build  an  equation  with  two  given  roots  twite  the  prod- 
net  of  X  less  the  first  root  and  x  less  the  second  root^  and 
set  this  product  equal  to  zero. 

Ex.  2.    Given  the  roots  2  +  V3  and  2  —  V3,  find  the  equation. 
The  equation  is         [.r  -  (2  +  ^'"5)]  [a;  -  (2  -  VS)]  =  0. 
Multiplying,  we  get  _^.o  ._  4  ^  ^  ^  ^  0, 

which  is  the  desired  equation.     (See  Example  1,  p.  207.) 

Ex.  3.    Given  the  roots  -|-3  and  —  \/2,  find  the  equation. 
The  equation  is  (x  — 3)  (x  +  V2)  =  0, 

or,  .r2  -  (3  -  V2)x  -  3  V2  =  0. 

The  coefficients  in  this  equation  are  surds ;  this  will  always  happeu 
if  there  are  siird  roots,  unless  the  example  is  especially  selected  with 
roots  of  the  form  a  +  Vh  and  (t  —  Vb. 

223 


224  QUADRATIC    EQUATIONS  [Cii.  VIII 

The  pi'inciple  of  tliis  article  may  als(j  be  stated  in  a  form 
called  the  factor  theorem      (See  also  Appendix,  §  5) : 

If  s  is  a  solution  of  the  equation  ax^  -\-  bx  -\-  c  =  0,  then 
(x  —  s)  is  a  factor  of  the  expression  ax"^  +  hx  +  c. 

118.    Relation  of  Roots  to  Coefficients.     If  r  and  s  are 

the  roots  of  a  quadratic  equation,  tlie  equation  is 

2)2  —  (r  +  s^x  +  rs  =  0, 
as  seen  above.     Hence, 

(1)  The  sum  of  the  roots  loith  the  sign  changed  is  the 
coefficient  of  x  in  the  equation. 

(2)  The  product  of  the  roots  is  the  last  {or  constant^ 
term. 

Notice  that  the  following  problem  arises  from  this :  Having  di- 
Anded  both  sides  by  the  coefficient  of  x%  any  quadratic  equation  has 
the  form  above;  to  find  its  roots  we  must  find  two  numbers  whose 
sum  and  product  we  know.  Problems  of  this  kind  lead  to  quadratic 
equations. 

Thus,  if  we  know  that  the  sum  of  two  numbers  is  6  and  their  prod- 
uct is  8,  these  numbers  must  be  the  two  solutions  of  the  equation 
x^-6  x+8  =  0,  i.e.2  and  4.  Another  method  will  be  found  later 
(See  Chapter  X,  p.  253.) 


EXERCISES 

VII:  CHAPTER    VIII 

Fin 

d  quadratic  equations 

whose  roots  are : 

1. 

2,  3.                         5. 

-3,-5. 

9. 

—  l^it 

2. 

3,  5.                         6. 

7,-9. 

10. 

6,  -6. 

3. 

3,  -5.                    7. 

3,i. 

11. 

o\     _  1  1 

wj,           -1 2- 

4. 

-  3,  5.                    8. 

4,  -H- 

12. 

-  3.4,  7.1. 

13. 

3+Vo,  3— V5. 

18. 

2  -  V3,  3 

-V3. 

14. 

_  4  -  Vrr,  -  4  +  ^/ 

''17.        19. 

a  +  b,  a  — 

h. 

15. 

3-V7,  3-I-V7. 

20. 

a,  a. 

16. 

-  6  +  V5,  2  -  V5. 

21. 

a,  —a. 

17. 

3  +  V5,  2-V5. 

22. 

a  -\-  Vi",  a 

-V6. 

117-1  lii]     PllOPKliriKS   OF   QUADRATIC    EQUATIONS    2:2r> 


State  the  sum  and  produot  of  the  roots  in  each  of  the  follow- 
ing equations ;  tlien  check  by  solving  and  actually  finding  the 
sum  and  product : 


23.  .r2-8.r+12  =  0. 

24.  x--3x  =  28. 

25.  it-- +  7  a- =  30. 

26.  of  +  4  a;  +  3  =  0. 

27.  2ar-.r-10  =  0. 


28.  6x^  +  x-i0  =  0. 

29.  ar  —  2px  +  jr  —  (/'  =  0. 

30.  4 a;^- 12a; +  9=0. 

31.  3a^  +  18x-10=0. 

32.  a^-6x  +  l  =  0. 


Find  two  numbers,  given  the  following  data : 


33 

34 

35 

36 

37 

38 

39 

40 

41 

Sum 

9 

7 

-3 

—  0 

10 

_  1 

6 

21 

Gjj 

0 

Product 

14 

-18 

-108 

6 

.3 

-n 

-12J 

9^ 

-4  7-' 

119.  Factoring  Quadratic  Expressions.  Another  result  of 
the  woik  just  done  is  a  new  method  of  factoring  quadratic 
eupressions,  i.e.  expressions  of  the  form  , 

ax^  -\-  bx  -\-  e, 
such  as  occur  on  the  left  side  of   a  quadratic  equation 
when  the  right  side  has  been  reduced  to  zero.     (For  ele- 
mentary method  see  §§  60,  61,  pp.  95-99.) 

Ex.  1.    Factor  the  quadratic  expression  2  x-  —  9  a;  +  4. 
2  a;2  -  9  X  +  4. 
Let  us  first  solve  the  quadratic  equation 

2x2-9x  +  4  =  0, 
which  is  formed  by  setting  the  given  expression  equal  to  zero. 
The  solutions  are, 

a;  =  4  and  x  =  ^  (see  p.  203). 
Hence,  (x  —  4)  and  (x  —  ^)  are  each  factors  of  2  x"^  —  9  x  +  4   (see 
§  117).     The  other  factor  is  2,  since  we  know  we  nuist  produce  2  x-  as 
the  first  term.     It  follows  that 

2  x2  -  9  X  +  4  =  2(x  -  4)(x  -  i), 
or,  2  x2  -  9  X  +  4  =  (x  -  4)(2  x  -  1). 


HEDRR'K   S    EI..    ALG. 


226  QUADRATIC    P:QUATI0NS  [Ch.  VIII 

The  desired  factors  are  therefore  x  —  4  and  2  x  —  1. 

Ex.  2.    Factor  x-  -  4  .v  + 1. 

The  solutions  of  x'^  -  4  x  +  1  =  0  are  x  =  2  +  ^/^^  and  x  =  2  -  VS; 

hence,  a-"-'  -  4  x  + 1  =  [x  -  (2  +  V':!)]  [x  -  (2  -  V:5)]  . 

The  desired  factors  are  (x  —  2  —  V:])  and  (x  —  2  +  \/;^j). 

It  is    to   be   noted   tliat   the   coefficients  in    such  factors    may  be 

irrational ;     thus,    2  +  Vs    is     irrational.  The    factors    are    strictly 

poli/nomiah,  however,   since   the  important  letters  do  not  occur  irra- 
tionally, but  only  in  simple  powers. 

Ex.  3.   Factor  x-  -  2. 

The  solutions  of  x-  -2  =  0  are  x  -  +  V2  and  x  =  —  V2; 
hence,  x^  -  2  =  [x  -  V2]  [x  -  (  -  V2)]. 

The  desired  factors  are  (x  -  V2)  and  (x  +  V2). 
Ex.  4.    Factor  x-  —  4  a;  +  4, 

The  only  solution  of  x'^  -  4  x  +  4  =  0  is  x  =  2.  Hence,  x^  -  4  x  +  4 
is  a  perfect  square  (x  —  2)-,  as  is  seen  on  inspection.  This  will  always 
happen  ifh'^  —  iac^O,  which  is  therefore  the  condition  that  ax'^  +  bx  +  c 
shall  be  a  perfect  square. 

Ex.  5.   Factor  ax-  +  hx  -\-  c. 


I)      "vh^  ■ —  4  (ic 

The    .solutions    of    ax'^  +  bx  +  c  =  0    are   x  = -— ±  — — ; 


lience, 


where  the  factor  a  on  the  outside  is  chosen  so  that  the  term  in  x-  will 
be  ax-. 

The  remit  is  meaningless  if  I?  —  4  a<;  <  0,  for  in  tliat  case 
the  expressed  radical  VP  —  ^ac  has  no  meaning  at  pre.s- j 
ent.       It    is    helpful    to    notice  the   following   scheme   of  j 
results : 


llP-li'o]     I'KOPKRTIES   OF   QUADRATIC   EQUATIONS    227 


lllM  ItlMlNANT 

SlU.ITrilNS    OK 

rta;2  +  ha-  +  c  =  (I 

Factors 
ax-  +  hx  +  c 

liositive 

real  unequal  roots 
(two  solutions) 

real  and  unequal 
(factorable) 

zero 

equal  roots 
(only  one  solution) 

equal :  perfect  square 

negative 

imaginary  roots 
(no  solution) 

imaginary 
(factoring  impossible) 

The  principle  to  be  used  here  is  nothing  more  than  that 
stated  in  §  117  :  If  r  is  a  root  of  a  quadratic  equation 
whose  right  side  is  zero,  Qx  —  r)  is  a  factor  of  the  left  side. 

120.    Solution  by  Factoring.     As  in  Ex.  5,  p.  226,  we 

may  factor  any  quadratic  expression,  the  factors  being 
meaningless  for  the  present  if  P  —  4  ac<0.  We  may  do 
this  same  work  indej^endently  as  follows: 

(4  aV  +  4abx  +  P)  -  (^2  _  4  ^c) 


ax^  +  bx-^  c  = 


4a 


as  is  seen  by  reducing  the  right  side.     Or, 


ax^  +  bx  +  c  = 


_(2ax  +  by  -  (  V^>2  -  4  acy 


4a 


in  which  form  the  factors  can  be  seen.     (See  also  Ex.  5, 
§  ^19-) 

Ex.  1.    a;2-4.T  +  l=(.'c2-4a;  +  4)-(4-l)  =  (x-2)2-(V3)- 
=  (.c  _  2  -  V3)  (a;  -  2  +  V3). 

As  in  this  example,  the  student  should  not  try  to  remem- 
ber the  formulas  above,  but  he  should  rather  try  to  com- 
plete the  square  as  in  §  206. 

It  is  interesting  to  notice  that  a  quadratic  equation  may 
always  be  solved  by  this  method  of  factoring : 

Since  x^-ix+\=(x  -2  +  V3)(x  -  2  -  V8), 


228  QUADRATIC    EQUATIONS  [Ch.  Till 

as  we  just  saw,  the  quadratic  equation 

3:2  -  4  ,r  +  1  =  0 
may  be  written  (x  —  2  +  V^)  (x  —  2  —  V^)  =  0. 

Hence,  either        (x  -  2  +  \/3)  =  0,  or,  (x  -  2  -  VS)  =  0. 

(See  §  66,  p.  107.) 

And  X  =  2  -  a/3,  or,  x  =  2  +  VS. 

This  method,  while  interesting,  is  not  often  used  in 
difficult  examples,  because  it  is  more  complicated  than 
the  methods  already  given.  In  easy  examples,  however, 
where  the  factors  may  be  found  readily  by  inspection,  the 
method  by  factoring  is  by  far  the  quickest.  We  have 
used  it  above  very  often ;  see  pp.   103-115,  151,  203. 

EXERCISES   VIII:    CHAPTER   VIII 

In  the  following  exercises,  note  the  values  of  a,  b,  c ;  de- 
termine b^  —  Aac  and  classify  the  expressions  according  as 
b'  —  4  ac  >  or  =  or  <  0.  When  b^  —  4  ac  ^  0,  find  the  roots  of 
the  corresponding  equations,  and  thus  factor  the  expressions : 

1.  ,^^_cfx  +  U.  8.  3  y-- 30  I' +  75.  15.   7?-2-9r  +  4. 

2.  2r-7^)  +  13.  9.  4«--20a-+25.  16.   2s--10s  +  12i. 

3.  6  m  — wr  — 9.  10.  7?r  — 12?(  +  4.  17.   II2-  + 2  2  —  40. 

4.  3f-^t-2.  11.  5«2-3?*  +  2.  18.  9^-  +  42<+49. 

5.  Z-  +  Z  +  1.  12.  5X--3.C-2.  19.   13t-  +  5t. 

6.  Z--Z  +  1.  13.  6c- -lie +  3.  20.  3t--2t  +  105. 

7.  z^  +  z-1.  14.  f--12t.  21.  a;2  4-.3.t-+.02. 

First,  factor  each  of  the  following  expressions,  then  find  the 
roots  of  the  corresponding  equation  : 

22.  a-  -  4.  26.   ay'  +  Sx-  33.  30.    6 2-  +  2  -  35. 

23.  4  a-  —  9.  27.   X-  +  5  X.  31.  4  nr  —  2  m  —  12. 

24.  25?;i2-l.  28.    10/-//.  32.    15r--34r  +  15. 

25.  (f-17g  +  30.  29.    ix".  33.   a;--.3x  +  .02. 


120-121]     PROPERTIES  OF   (iUADRATIC    EQUATIONS     229 

121.  Literal  Coefficients.  The  student  has  solved  some 
exercises  in  which  letters  occurred  in  the  coefficients.  We 
shall  now  solve  such  equations  systematically. 

Ex.  1.    Given  ar  —  2  mx  +(m-  —  1)=  0. 

Transpose  ?n''  —  1  and  add  m'^  to  both  sides  : 

x^  —  2  mx  +  ?«2  =  —  {iii^  —  I )  +  '"^ 
or,  (.c  —  my-  —  \,         or,  x  —  m  =±\,         or,  x  =  m  ±\. 

This  solution  holds  for  any  value  of  the  letter  m.     (Check  it.) 

Ex.  2.    Given  x^  - 12  xy  +  4  i/-  =  0. 

Solving  for  x,  we  find  a;^  —  12  x//  4  36  y-  =  32  y% 
or,  X  —  Q  y  —  ±  V32  y'^, 

or,  x  =  6?/±\/32p=?/(G±4V2).     (Check  it.) 

Solving  for  y,  we  find  4  »/-  -  12  x^  +  9  x^  =  8  x^, 
or,  2  ^  -  3  X  =  ±  y/%7\ 

or,  2/=^±^V8T^  =  ?(6  iiVS).     (Check  it.) 

Thus,  we  may  solve  the  same  equation  for  any  letter  in  it. 

*  Note  on  the  Discriminant.  The  given  equation  has 
equal  roots  if  the  discriminant  is  zero. 

Let  (I  stand  for  discriminant ;  then  the  roots  of  the  given  equa- 
tion are  real  and  unequal  if  (?  >  0;  imaginary  if  d  <  0;  equal 
if  d  =  0. 

Ex.  3.    Given  x'-  +  l\v  +  (3  +  k)  =  0. 
Comparing  with         ax-  +  bx  +  c  =  0, 
we  find  a  =  1,  b  =  L',  c  =  3  4  /,•;  hence,  the  discriminant 
b-'  -  4  rtc  =  k^  -  4  (3  +  k)  =  k'-^ik-  12. 
Let  us  try  several  values  of  k\     If  /;  =  0,  r/  =  -  12,  and  the  roots 
are  imaginary ;  in  fact,  the  equation  is 

a,.2  +  0  •  X  +  (3  +  0)  =  0, 
or,  x2  +  3  =  0, 

or.  X  =  ±  V  -  3,  which  is  imaginary. 

*  This  work  may  be  omitted  unless  especially  desired  ;  in  any  case  it 
should  not  be  attempted  until  the  student  is  quite  proficient. 


230 


QUADRATIC    EQUATIONS 


[Ch.  VIII 


Draw  the  figure  for  k  =  0,  and  show  that  it  does  not  cut  the  main 
horizontal  line. 

U  k  ~l,  f/  =  1  —  4  —  12  =  —  15;  roots  imaginary. 

What  is  tlie  origiiial  equation  in  tliis  case  (k  =  1)? 

Solve  it.     Are  the  solutions  imaginary  ?     Draw  tlie  figui-e. 

If  y^-  =  6,  J  =  36  -  4  .  6  -  12  =  0;  roots  equal. 

What  is  the  original  equation  in  this  case  (k  =  6)  ? 

Solve  it.     Are  the  roots  equal  {i.e.  only  one  solution).     Draw. 

If  k  =  10,  (^  =  100  -  4  •  10  -  12  =  48 ;  roots  real  and  unequal. 

What  is  the  original  equation  in  this  case  (^k  =  10)  ? 

Solve  it.     Are  the  roots  real  and  unequal?     Draw. 

Trying  several  other  values  of  k,  we  make  this  table  : 


k 

0 

1 

2 

.S 

etc. 

5 
—  7 

(J 

7 
9 

etc. 

-  1 

2 

-:) 

etc. 

d 

-12 

-15 

0 

-7 

0 

9 

roots 

imag. 

imag. 

imag. 

equal 

real 

imag. 

equal 

imag. 

[Let    the    student    fill    in    the    blank    spaces    and    extend    this 
table  to  k—-\-  10  and  backward  to  k  =  —  10.] 

We  may  draw  tliese  values  of 
k  and  d  as  shown  in  Fig.  40. 
From  this  it  is  clear  that : 

r/  =  0  only  tchen  k  =  —  2  and 
ujIioi  k  =  Q  (i.e.  the  original 
equation  has  equal  roots  onii/  when 
^•=  -lG  or  -  2). 

</ <  0  ivhen  k  has  any  value 
helwecn  —  2  aii'l  +  0  (I.e.  the 
original  equation  has  imaginary 
roots  when  and  only  when  k  is  a 
numl)er  between  —  2  and  +  G). 

r/ >  0  irhen  k  is  less  than  —2, 
also  when  k  is  f/reatrr  than  +  0 
(i.e.  tlie  original  eciuation  has  real 
and  unecjual  roots  when  k<C  —  2, 
and  when  k>  +  (i). 

These  results  may  also  he  found 
hy  factoring  the  discriminant: 
Fig.  40.  rf  =  ^2  -  4  k  -  12  =  (k  +  2)  (k  -  U). 


iL'l]         PROPERTIES    OF    Ql'ADRATIC    EQUATIONS        281 

EXERCISES    IX:    CHAPTER   VIII 

Solve  the  following  equations  for  the  letters  indicated: 

1.  x^  —  o  ax  =  —  G  cr.     (First  for  x ;  then  for  a.) 

2.  X-  —  G  axr  =  —  5  a^.     (x,  a.) 

3.  vr  —  G  vix  =  7-  —  9  ar.     (m,  ic.) 

4.  mr  —  (vi  +  n)rs  +  ?is^  =  0.     (?•,  s.) 

5.  2  z-  —  3  mz  —  14  7«-  =  0.     (2,  7*1.) 

6.  3  a-  +  8  ah  -  3  ?>-.     («,  b.) 

7.  ^r_2^,m-(2m  +  l)  =  0.      (jj.) 

8.  ?/'-  —  G  a//  —  6  a  —  1  =  0.     (?/.) 

9.  «-—  mt  +  m  =  l.     (t.) 

10.  or  +  G  y-  +  G  z-  +  12  yz  —  5zx  —  5xy  =  0.     (a;,  ?/.  z.) 

Find  for  what  values  of  k  the  roots  of  the  following  equa- 
tions will  be  real  and  iineqnal,  real  and  eqnal,  or  imaginary  ; 
in  the  first  two  cases,  solve  the  equations. 

[Omit  these  until  very  proficient.] 

11.  .r- —  A-.r  +  A- =  0. 

12.  x^-{k  +  2)x  +  2(k  +  2)=0. 

13.  x-  +  2(k-3)x  +  {k  +  3)=0. 

14.  A-.T;-  +  {k  +  5)  x  +  (k  +  5)  =  0. 

15.  ^^  +  ^'  +  ——=0. 
]^  +  'S     h     k  +  3 

16.  The  area  of  a  square  field  measured  in  square  rods  is 
equal  to  the  length  of  its  whole  perimeter  measured  in  rods 
less  twice  a  given  number  A:.  For  what  values  of  k  is  the 
problem  possible  ?  impossible  ?  For  what  values  are  there 
two  answers  for  the  size  of  the  field  ?  When  is  there  just 
one  answer  ? 


232  QUADRATIC   EQUATIONS  [Cn.  VII] 

REVIEW   EXERCISES   X:    CHAPTER   VIII 

Solve  the  following  equations  : 

1.  z--5z-'S00  =  0.  11.  12ar'-a;-20  =  0.  j 

2.  ^^_  3^-108  =  0.  12.  18  cr _  79  a _  3100  =  0. 

3.  ir  +  34 X- 800  =  0.  13.  4k^  +  7k-U7  =  0. 

4.  a-2 -  29  a;  + 168  =  0.  14.  13  A-^4- A;- 120  =  0. 

5.  2>'-6i>- 247  =  0.  15.  28  x" -3  X -135  =  0. 

6.  6:i---7«-20  =  0.  16.  2y--oy-S8  =  0. 

7.  15  7-2 -  22  r- 91  =  0.  17.  18/-31?/ +  6  =  0. 

8.  6g2  +  5r/-781  =  0.  18.    6:1-2+7.^-49  =  0. 

9.  14  2/-- 3  2/ -270  =  0.         19.    12ar  +  5  .«- 72  =  0. 
10.    8  6^- 18  s -425  =  0.  20.    35A;--^•-6  =  0. 

Factor  the  following  expressions  by  first  solving  the  corre- 
sponding equations : 

21.  6ar'-x-77.  26.  21r  +  13r  +  20. 

22.  24jy--94j>-63.  27.  6Z--7  2-24. 

23.  ()k--7Jc-33.  28.  50 /•- -  5 /•  -  36. 

24.  ?/-'  -  66  11/  -  675.  29.  21  m-  -  20  m  -  96. 

25.  12  A-  +  A-  -  130.  30.  24  X-  + 19  x  -  35. 

Form  equations  whose  roots  shall  be : 

2—3  —1—2  r-- 2 

31.  6,-8.       33.    3,^.         35.    ^,^.        37.     .,^. 

32.  i   i  34.    5,:^.         36.    3,1  38.     -5,11 

39.   a- 3 6,2a- i.  40.    ?, -• 


Ijl]  RKVIKW  233 

Find  the  discriminant  of  each  of  the  following  equations; 
determine  the  character  of  the  routs,  and  solve  if  the  roots  are 
real : 

41.  14  .)•■-  + 20  X-- 15  =  0.         45.    .3.r-'-11.7;+ll  =  0. 

42.  81  Z--1  OS  2  +  121=0.      46.    6.T--ll.c  +  4  =  0. 

43.  St--VU  +  6  =  0.  47.    x--{Jc  +  3)x  +  Jc  =  0. 

44.  oy-  +  U)y  +  ll  =  0.  48.    t- -  (k  +  3)t- k  =^  0. 

49.  (k-l)r  +  2kr+(k+l)=0. 

50.  (A-  +  1)  .r  -  (.".  k  +  1)  X  +  (A-  +  1)  =  0. 

51.  The  sum  of  two  numbers  is  16;  the  sum  of  their  squares 
is  loO.     What  are  the  numbers  ? 

52.  The  sum  of  two  numbers  is  s;  the  sum  of  their  squares 
is  .s.  Show  that  the  quadratic  equation  found  in  order  to 
determine  one  number  has  for  its  other  root  the  other  number. 

Sketch  ok  Solution.  Let  n^  and  n.,  denote  the  two  iiuinbers. 
Then  n^  +  n2  =  s  or  n^  =  .s  -  n.^ ;  since  «,-  +  n.^  =  s,  the  equation  for 
either  n  is  of  tlie  form  n^  +  (s  -  ny  =  s ;  hence  the  two  n's  are 
precisely  the  two  roots  of  this  equation. 

53.  The  sum  of  two  numbers  is  s ;  if  each  of  the  numbers  is 
divided  into  .1,  the  sum  of  the  quotients  is  s.  Show  that  the 
same  quadratic  equation  has  both  numbers  for  its  roots. 

54.  Four  consecutive  integers  have  as  the  sum  of  their 
squares  54.     What  are  the  integers  ? 

55.  A  and  B  together  can  fold  1000  circulars  in  an  hour. 
It  is  observed  that  when  each  works  separately  at  1000  circu- 
lars, A  finishes  50  minutes  before  B.  How  many  circulars 
can  each  man  fold  in  an  hour  ? 

56.  Two  tanks  have  capacities  in  the  ratio  3  to  4.  W'hen 
unequal  pipes  are  attached,  the  small  tank  is  tilled  in  2 
hours  less  time  than  the  large  tank.  The  two  tanks  are  then 
connected,  and  both  pipes  \ised  to  empty  them.  This  process 
requires  2  hours  48  minutes.  How  long  would  be  required 
for  each  pipe  alone  to  empty  each  tank  ? 


234  QUADRATIC    EQUATIONS  [Ch.  VIII 

[Suggestion.  The  student  may  introduce  the  idea  of  a  "  unit 
tank  "  whose  capacity  is  one  third  that  of  the  smaller  tank ;  use  as 
the  principal  unknown  the  time,  /,  required  to  fill  the  small  tank  with 
its  pipe,  and  express  the  capacity  of  each  pipe  (Cj  and  c,),  i.e.  the 
amount  each  can  carry  in  one  hour,  in  terms  of  ^] 

57.  An  open  box  8  cm.  high,  whose  base  is  a  rectangle  with 
sides  in  the  ratio  3  to  4,  is  to  have  a  surface  area  of  640  sq.  cm. 
What  are  the  dimensions  of  the  base  ? 

58.  A  regular  pyramid  on  a  square 
base  is  to  be  made  by  folding  a  figure 
like  that  sliown.  The  altitude  of  the 
triangular  sides  is  to  be  6  in.  The  sur- 
face area  of  the  pyramid  is  7iot  to  exceed 
100  sq.  in.  What  must  be  the  side  of 
the  base  to  obtain  precisely  that  surface 
area?  (Solve  first  graphically  and  then 
Fig.  41.  directly  from  the  equation.) 

59.  What  three  consecutive  integers  can  measure  the  sides 
of  a  right  triangle  ? 

60.  Show  that  if  one  perpendicular  side  and  the  hypotenuse 
of  a  right  triangle  are  measured  by  consecutive  integers,  the 
square  of  the  other  perpendicular  side  must  be  measured  by 
an  odd  number.  Choose  for  the  square  of  this  side  successively 
9,  25,  49.     In  each  case  what  are  the  other  two  sides  ? 

61.  A  body  is  thrown  upward  at  a  height  of  500  feet  with  a 
speed  of  30  feet  a  second.  When  will  it  reach  the  earth  ? 
(Solve  first  graphically,  then  directly  from  the  equation.) 
See  p.  221. 

A  body  thrown  horizontally  into  the  air  at  a  height  of  h  feet  with 
a  speed  v  feet  a  second,  follows  a  path  thus  described :  in  t  seconds 
the  horizontal  flight,  x  feet,  and  the  distance  from  the  earth,  y  feet,  of 
the  body  are  given  by  x  =  vt,  y  =  h  —  Ifi  t'^.  A  body  thrown  into  the 
air  from  the   earth's  surface  at  an  angle  of  45°  with  a  speed  of  v 

foUowa  the  path  x  =  -^,  y  =  ^  -  16  <=*• 
^2  V2 


\-jl]  REVIEW  285 

62.  What  is  tlie  relation  between  x  and  y,  i.e.  the  equation  of 
the  path  ofthebodi/,  in  each  of  the  above  cases? 

63.  A  stone  is  thrown  horizontally  from  a  cliff  400  feet  high 
at  a  s{)eed  of  50  feet  a  second.  When  will  the  stone  strike 
the  earth,  and  how  far  from  the  foot  of  the  cliff  ? 

64.  When  will  the  horizontal  distance  traveled  by  the  stone 
be  equal  to  its  height?  What  will  this  distance  be?  (Solve 
graphically  first.) 

65.  A  body  is  thrown  into  the  air  at  an  angle  of  45°  with 
a  speed  of  37  ft.  a  second.  At  what  horizontal  distance  will 
it  be  5  ft.  from  the  ground  ?     How  long  has  it  then  traveled  ? 

66.  At  what  horizontal  distance  will  the  body  be  feet 

128 
from  the  ground  ? 

Consider  the  graph  carefully  in  connection  with  Exs.  65,  (Uj. 

[The  following  exercises  are  intended  only  for  use  upon  a  review 
of  the  whole  book  and  are  not  to  be  solved  until  the  student  has 
completed  the  .study  of  Chapter  XII.] 

Solve  the  equations : 

67.  ..^+G..-^  =  40.  ,^     V^T7+       ^       =311 

68.  V.«  +  5A/.t;  =  14.  Va;  +  7 

69.  (.'c2  +  l)-4-8(x'^  +  l)  =  180.  72.    ar^-3a;-3Va!--3a,-+7  =  3 

70.  ■V2x  +  :i  +  Gx  =  71.  73.    ^/\v-{-i/x  =  2. 


74.  2  .r  +  7  .r  +  6  -  3  V8  x-  +  28  .r  -  1 1  =  0. 

75.  2  (x  +  ^Y'-  9  fx  +  IV  10  =  0. 


236  QUADRATIC   EQUATIONS 


SUMMARY   OF   CHAPTER   VIII:  QUADRATIC   EQUATIONS, 

pp.  203-235 

Part  I.     Methods  of  Solution,  Character  of  Roots, 

pp.  203-215. 

Definition  of  Quadratic  Equations  :  contains  square  of  unknown. 

First  Methods  of  Solution :  factoring,  as  in  Chapter  IV,  if  easy; 
otherwise,  drawing  figure.     Exercises  I.         §  108,  pp.  203-205. 

General  Solution,  Completion  of  Square:  typical  example;  definition 
of  root;  insistence  on  verification.  §  109,  pp.  205-206. 

Formal  Rule  for  Completion  of  Square :  making  left-hand  side  perfect 
square  by  adding  a  number.     Exercises  II.    §  110,  pp.  200-208. 

Second  Method:  multiplication  by  4  x  1st  coefficient;  previous 
method  witli  kx  in  place  of  x.    Exercises  III.    §  1 1 1,  pp.  208-209. 

Equal  Roots :  single  answer;  perfect  square;  curve  "tangent "  to 
main  horizontal  line.  §  112,  pp.  209-210. 

Imaginary  Roots :  no  answer;  curve  misses  main  horizontal;  defi- 
nition of  imaginaries.     Exercises  IV.  §  113,  pp.  210-212. 

Formula :  general  solution  ax'^  +  bx  +  c  =  0.       §  111,  pp.  212-214. 

Discriminant:  correspondence  of  h'^  —  iac^O  to  2  ("real  and 
unequal ")  roots,  =  0  to  1  ("  equal  ")  root,  <  0  to  0  ("  imagi- 
nary ")  root ;  irrational  roots.    Exercises  V.    §  115,  pp.  214-215. 

Part  II.     Practical  Applications;  Problems,  pp.  216-222. 

Practical  Examples  :  solution  ;  suggestions.     Exercises  VI. 

§  lie,  pp.  216-222. 

Part  III.     Properties  of  Quadratic  Equations,  pp.  223-235. 
Given  Roots:  correspondence  of  roots  r  and  s  to  equation 

(x  —  r)  (x  —  s)  =  0. 
Factor  Theorem :  discovery  of  factor  x  —  r  for  any  root  r. 

§  117,  pp.  223-224. 
Relation  of  Roots  to  Coefficients:  sum  of  roots  -  —  coefficient  of  x; 

product  =  constant  term.     Exercises  VII.      §  118,  pp.  224-225. 
Factoring  Quadratic  Exj)ressions :  use  of  factor  theorem;  nature  of 

factors  from  discriminant.  §  119,  pp.  22.5-227. 

Solution  by  Factoring :  factors  of  ax- -\- bx -[- c ;    method  advisiable 

only  in  simple  cases.     Exercises  VIII.  §  120,  pp.  227-228. 

Literal  Coefficients :    typical  problems;    discriminant;    figure    for 

discriminant.     Exercises  IX.  §  121,  pp.  229-231. 

Review  Exercises  for  Chapter  VIII :  Exercises  X.  pp.  232-235. 


CHAPTER   IX 

VARIATION:    INDETERMINATE    EQUATIONS 

122.    Simple  Variation.     We  have  discussed  before  quan- 
tities tiiat  vary.      Thus,  the  cost  of  an  amount  of  butter 
varies  with  the  nuniljer  of  jxninds  bought.      (See  p.  23.) 
I      Whenever  the  quotient  of  tivo  varyimj  quantities  y  and  x 
is  a  constant  k, 

^  =  k,  or,  y  =  kx, 

the  variables  y  and  x  are  said  to  be  in  proportion  (see  pp. 
25,  140,  etc.),  by  which  we  mean  that  Any  pair  of  values 
of  y  and  x  form  a  proportion  witli  any  other  pair.  We 
also  say  in  this  case  that/  varies  directly  as  x. 

We  have  seen,  p.  141,  that  the  corresponding  figure  is  a 
straight  line  through  the  starting  point. 

Tlie  following  are  therefore  synonymous : 

(1)  y  is  proportional  to  x. 

(2)  y  varies  directly  as  x. 

(3)  The  quotient  y  -^x  is  constant ;  or,  y  =  kx. 

(4)  The  graph  for  y  and  x  is  a  straight  line  through  the 
starting  point. 

Instead  of  the  quotient  y  -^  x,  we  may  speak  of  the  ratio 
of  y  to  X  and  write  it  y  :  x  ;  and  we  may  say  that  the  ratio 
y  :  a;  is  a  constant.  This  constant,  which  is  called  k  above, 
is  called  the  ratio  of  proportionality. 

Thus,  if  butter  costs  30  (*  per  pound  (see  p.  28),  the  ratio  c:p  or 
c 

~,  where  c  denotes  the  cost  in  cents  and  p  denotes  the  number  of 
P 

237 


238    VARIATION:   INDETERMINATE   EQUATIONS    [Cn.  IX 

pounds,  is  always  30.  The  ratio  of  proiiortionality  is  30.  Any  pair 
of  values  of  c  and  j),  say  Cj  and  p^,  give  the  same  ratio  as  any  other 
pair,  Cg  and  p^ : 


C-, 


1^ 


(=30),  or,  ^  =  £i.     (See  VI,  (1),  p.  138.) 


Pi     Ih  ^2      P2 

If  one  varying  quantity,  2,  varies  directl}^  as  the  prod- 
uct, xy.y^  of  two  other  quantities  x  and  y^  then  z  evidently 
varies  directly  as  x  when  y  is  constant,  and  as  y  when  x  is 
constant.  For  if  z  =  k  ■  x  •  y^  where  k  is  a  constant,  tlien 
z  =  (k  •  x^  •  y^  whence  2  is  a  constant  times  y  \i  a;  is  a  con- 
stant. Likewise  z  =  {k  •  y')  •  x,  whence  2  is  a  constant 
times  2:  if  ?/  is  a  constant. 

Thus,  the  area,  ^4,  of  a  rectangle  is  given  by  the  formula 
A  =b  x  h, 
where  h  is  the  base  and  h  is  the  height  of  the  rectangle,  measured  in 
feet  or  in  any  other  unit  of  length.  If  h  is  constant,  A  varies  as  h, 
i.e.  the  areas  of  rectangles  with  equal  bases  are  to  each  other  as  the 
heights.  Likewise  A  varies  as  6  if  ^  is  constant,  i.e.  the  areas  of  rec- 
tangles of  the  same  height  are  to  each  other  as  the  bases. 

The  reverse  statement  is  also  true :  if  z  varies  as  x  when 
y  is  constant,  and  as  y  when  x  is  constant,  then  z  varies  as 
the  product  x  x  y. 

Thus,  if  we  know  that  the  area,  ^,  of  a  rectangle  varies  as  h  (the 
height)  when  b  (the  base)  is  a  constant,  and  as  h  when  A  is  a  constant, 
we  may  conclude  that  A  varies  as  ft  x  /;.  A  formal  proof  of  this  last 
form  of  statement  is  deferred. 

123.  Linear  Variation.  We  have  also  seen  that  a  linear 
equation  (or  equation  of  the  first  degree)  of  the  form 

(1)  y  =  ax  +  b, 

where  a  and  h  are  constants,  is  represented  by  a  straight- 
line  graph.  CSee  pp.  2P>,  143.)  This  kind  of  relation  is 
often  called  linear  variation,  and  we  say  that  ?/  is  a  linear 
function  of  x.     Examples  of  this  occur  on  pp.  25,  143. 


li'L'-lL>4]   VARIATION:    IXDETEKMINATE    EQUATIONS    239 


Ex.  1.    As  another  example  consider  the  equation 

2  .K  -  3  //  =  G. 
Solve  for  ^ :  y  =  lx  -2, 

■which  is  of  this  same  type,  i.e.  the  curve  is  a  straight  line.  To  draw 
it  we  need  only  plot  two  points  (see  footnote,  p.  1-44).  For  example, 
(x  =  0,  y  =  -  2)  and  (^  =  0,  a;  =  3)  ^ 

as  can  be  seen  also  from  the  origi- 
nal equation  2  x  -  3  y  =  6.  The 
graph  is  as  shown  in  Fig.  -42. 

Any  equation  of  the  form 
(2)  Ax  +  By  ^C  =  0 

can  be  reduced  to  the  type  (1)  pro-  * 
vided  B  ^  0,   as  in   the  preceding 
example.     If  B  =  0,  the  equation 


1  1  ;  '  I  [  !  r  T 

,   V   1^ 

-iV  k  ul  -i-X^ 

,:....:9;:"Js'' ,   "     -  : 

.'I        1 

^^              1 , 

^ 

■ ..'                 ' 

-V                             1 

.. 1    1 

Ax  +  C  =  0,  or,  X  = 


Fig.  42. 


r/o(?.«  no<  contain  y.     Hence,   x  =  the  same  thinrj  for  every  possible  y, 
and  the  curve  is  a  vertical  straight  line. 

In  all  cases,  unfJiout  exception,  an  equation  of  the  first 
degree  is  represented  in  a  figure  hy  a  straight  line. 

124.  Inverse  Variation.  It  may  happen  that  one  vary- 
ing quantity  increases  as  another  decreases,  in  such  a  way 
that  their  product  is  constant. 

Thus,  if  a  train  goes  20  miles  per  hour  on  a  trip  600  miles  long,  the 
time  taken  is  30  hours ;  if  it  goes  25  miles  i^er  hour,  the  time  taken 
is  24  hours ;  if  it  goes  30  miles  per  hour,  the  time  taken  is  20  hours. 
Notice  the  product  speed  x  time  =  constant  =  total  distance : 

,s  •  <  =  d, 

where  s  stands  for  speed,  t  for  time,  d  for  distance.     A  table  follows : 


s 

1 

2 

5 

10 

15 

20 

30 

40 

50 

60 

etc. 

-  10 

-30 

etc. 
etc. 

t 
d 

600 
600 

300 
600 

120 

60 

40 

30 

20 

1.5 

12 

10 

etc. 

-60 

-20 

600  I  600 

600 

600 

600 

600 

600 

600 

600 

600 

240    VARIATION:   INDETERMINATE  EQUATIONS    [Cii.  IX 

Here  d  is  a  constant,  600  (in  miles).  But  s  and  /  vary.  The 
graph  of  the  relation  between  s  and  /  is  shown  in  Fig.  i'i.  Negative 
values  of  *•  and  t  correspond  to  backward  motion. 

A  relation  between  two  varying  quantities  —  say  y  and 

X  —  such  that  f  h  Tc 

X  •/  =  k  [or  x=  -,  or  y  =  - 

V  ^  X 

where  k  is  a  constant,  is  called  inverse  variation,  and  we 
say  that/  varies  inversely  as  x. 


~]~ 

(_  1 

i 

1 

i:                    c 

V 

H-                      \ 

in                "^^ 

\                                             1 

-\                      6(-=6CI0 

Jt     i 

^-iCl 

"    "            -H-       -^   ' 

I             , ,_ 

Fig.  43, 

Curves  shaped  like  the  above,  i.e.  curves  corresponding  to  an  equa- 
tion of  the  form  xy  =  k\  we  shall  call  incense  variation  curves.  Simi- 
larly, we  notice  that  direct  variation  curves  are  straight  lines. 

EXERCISES   I:     CHAPTER   IX 

Draw  graphs  to  represent  the  following  relations : 

1.  y  =  lGx.  5.    3x  +  7?/  =  — 18. 

2.  ij  =  — 12 X +  7.  6.    y^o{x  —  2). 

3.  x  =  3y  —  5.  7.    xy  =  1. 

4.  2,T  +  r>.7/  =  9.  8.    xy  =  100. 


l_'l]       VARIATION:    INDKTKII.MINATE    EQUATIONS        241 

9.  xy  =  72.  13.  a;?/ =  —  600. 

10.  .r//  =  540.  14.  x(y-'S)=  100. 

11.  x}/  =  —  l.  15.  {x  —  2)y  =  75. 

12.  .r^  =  -100.  16.  (.r- 3)0/ -2)  =  500. 

17.  Suppose  y  varies  tlii-ectly  as  x  and  the  ratio  of  propor- 
tionality is  2  ;  write  the  equation  ;  draw  a  graph  ;  find  y  when 
X  =  1,  when  x  =  2. 

18.  Suppose  y  varies  directly  as  x  and  x=2  when  y  =  10; 
write  the  equation;  draw  a  graph;  find  y  when  a;  =  3, 

19.  Suppose  y  varies  inversely  as  x  and  x  =  l  when  y  =  2; 
write  the  equation;  draw  the  graph;  find  y  when  x  =  2. 

20.  Suppose  y  varies  as  the  square  of  x  and  x  =  l  when 
y  =  4 ;  write  the  equation  ;  draw  the  graph  ;  find  y  when  x  =  2, 
when  X  =  3, 

21.  The  volume,  V,  of  a  box  whose  height  is  4  ft.  is  V=  4  id, 
where  iv  is  the  width  and  I  is  the  length  (in  feet),  and  where  V 
is  the  volume  (in  cubic  feet).  Show  that  V  varies  as  to  when  I 
is  constant,  and  as  I  when  w  is  constant. 

22.  Show  that  the  area  of  an  open  circular  cylinder  varies 
as  the  height  {h)  when  the  radius  (r)  of  the  base  is  a  constant, 
and  as  /•  when  h  is  constant.     (See  Tables.) 

23.  Find  from  the  Tables  all  geometrical  figures  whose  areas, 
or  volumes,  vary  directly  as  certain  of  their  dimensions,  and 
express  each  of  these  both  by  formulas  and  in  words. 

24.  The  cost  of  any  number  of  poiinds  of  butter  varies  as 
the  number  of  pounds.  If  one  pound  costs  30  cents,  express 
the  cost  of  any  number  of  pounds ;  draw  the  graph ;  what  is 
the  cost  of  5  pounds  ? 

25.  For  a  certain  mass  of  gas,  it  is  observed  that  pressure 
times  volume  is  equal  to  120,000.    Plot  the  graph.  (See  p.  218.) 

26.  Indicate  by  a  picture  the  relation  between  base  and  alti- 
tude of  a  triangle  of  constant  area. 

heurick's  el.  alg,  —  10 


242   VARIATION:   INDETERMINATE   EQUATIONS    [Ch.  IX 

27.  If  a  certain  stretched  wire  is  caused  to  vibrate,  then 
the  number  n  of  vibrations  a  second,  the  radius  r  of  the  wire 
in  centimeters,  and  its  length  I  in  centimeters  are  observed  to 
satisfy  the  relation  7irl  =  16000. 

For  a  wire  50  cm.  long,  plot  the  relation  between  n  and  r. 

28.  Express  by  a  figure  the  radius  and  length  of  all  wires 
sounding  i^of  the  middle  register  (n  =  320  per  second). 

29.  The  surface  area  of  a  cylindrical  ring  is  330  sq.  in. 
Express  by  an  equation,  and  plot  a  figure  for  all  possible 
values  of  the  radius  and  height  of,  the  ring.     (Take  tt  =  ^7-^-.) 

30.  The  relation  between  the  mass,  volume,  and  density  of 
a  body  is  m  =  vcl. 

Choose  a  suitable  constant  value  for  the  density,  and  plot 
the  relation  between  volume  and  mass. 

31.  Choose  a  suitable  constant  value  for  the  mass,  and  plot 
the  relation  between  volume  and  density. 

32.  Express  graphically  the  relation  between  the  total  sur- 
face area  and  the  slant  height  of  a  right  circular  cone  whose 
base  has  a  radius  of  7  inches. 

125.  Indeterminate  Equations.  The  examples  above  are 
all  examples  of  indeterminate  equations,  i.e.  equations  in 
which  each  of  the  unknown  letters  has  not  one  fixed  value, 
but  rather  an  indefinite  number  of  possible  values. 

Sometimes  a  problem  is  of  such  a  nature  that  only  a 
few  possibilities  remain ;  we  may  then  fix  definite  values 
for  the  letters. 

Ex.  1.  A  carpenter  has  boards  8  in.  wide  and  others  6  in. 
wide.  How  many  of  each  may  he  take  to  make  a  walk  50  in. 
wide,  the  boards  being  laid  lengthwise  ? 

Let  X  —  number  of  8-iiich  boards,  and  y  =  number  of  6-inch  boards. 

Then  8x  +  6y  =  width, 

or,  8  X  +  6  y  =  50. 

Let  us  now  try  various  luimbfirs.  If  a:  =  0,  y  —  ^/  =  8|.  (A  iu 
Fig.)     If  //  =  0,  x  =  ^^^-  =  (5^.     (B  in  Fi-.) 


li'4-12.-)]    VARIATION:  INDETERMINATE    EQUATIONS   243 
Y 


— 

— 

— N 

s. 

\ 

\ 

S, 

s. 

\ 

s 

10 

\ 

^ 

V 

\ 

s. 

\j 

K 

s 

\ 

V 

s 

[ 

1 

0 

B\ 

\ 

- 

\ 

\ 

s 

S 

- 

\, 

\ 

1 

1 

' 

1 

Fig  44. 

The  figure  is  a  straight  line,  for  the  equation  is  of  the  first  degree. 
Hence,  the  possible  values  of  x  and  y  correspond  to  the  points  on  the 
straight  line  joining  A  and  B. 

Now  the  carpenter  must  choose  an  integral  number  of  each  sort  of 
boards.  The  values  of  x  and  y  must  then  be  integers,  i.e.  the  corre- 
sponding points  lie  at  the  intersections  of  the  square  paper.  We 
notice  three  points  that  may  serve  : 

X  =  1,  1  [  ^  =  -*  1  f  a;  =  4, 

..y  =  7,|  I?/ =  6,  J  U  =  3, 

for  each  of  these  lies  at  least  very  close  to  the  straight  line.  Trying 
these,  we  find  x  =  2,  //  =  6  gives  a  total  width  2  •  8  +  6  •  6  =  .52  in. 
This  is  wrong. 

The  point  (x  =  1,  y  =  7)  gives  1  •  8  +  7  ■  6  =  50.  This  is  correct. 
Likewise  (x  =  4,  y  =  3)  is  correct.  The  carpenter  may,  therefore, 
choose  1  board  8  in.  wide  and  7  boards  6  in.  wiile,  or  4  boards  8  in. 
wide  and  3  boards  6  in.  wide.  If  he  has  more  8-inch  boards  than 
6-inch  boards,  he  should  clearly  choose  the  latter  scheme. 


244    VARIATION:   INDETERMINATP:   EQUATIONS  [Ch.  IX 

Ex.  2.  A  man  agrees  to  exchange  young  hogs  at  $  3.00 
each  for  sheep  at  $  5.00  each.  How  many  sheep  and  hogs 
may  be  traded  practically  ? 


~ 

~ 

~~ 

~ 

~- 

— 

S 

"0 

y 

• 

^ 

>i 

y 

y 

y 

y 

y 

y 

1 

Q 

y 

^ 

y 

y\ 

y 

B 

y 

y 

y' 

y 

0 

i' 

1 

0 

30 

h 

y' 

A 

y 

^ 

^ 

Fig.  45. 


Let  h. 
Then 


tlie  nninber  of  hogs,    and  s  =  the  number  of  sheep. 

3  /*  -  5  s  =  0. 
If  ^  =  0,   5  =  0.     If  li  =  10,   s  =  6. 

The  graph  is  surely  a  straight  line  since  the  equation  3^  —  5.'?=  0, 
linear.      Drawing  a  figure,  the  possibilities  are  seen   to  be  only 


these 


Hogs      .... 

0 

5 

10 

15 

20 

et€. 

Sheep     .... 

0 

3 

6 

9 

12 

etc. 

Ex.  3.  A  frame  (box  without  a  bottom  or  top)  2  ft.  high  is 
to  be  constructed  out  of  a  board  1  ft.  wide  and  24  ft.  long. 
Find  how  to  cut  the  board  to  make  the  largest  box. 


125]       VARIATION:    LNDETEKMINATE   P:QL'AT10NS      .245 


Let  s  =  length  of  one  side, 

then  6  —  s  =  length  of  other  side. 

Then  area  of  bottom  =  j*  (6  —  s) . 
G-s  Tlie  largest  box  is  that  which  has   the 

largest  bolloui.  We  wish  to  find  when 
h  =  *■  (6  —  .s)  is  largest,  where  b  means  the 
area  of  the  bottom  in  square  feet.  If 
s  =  0,A  =  0;    iis  =  l,b  =  5;iis^2,b=S; 


etc. 


Fui.  4i;. 
Proceeding   in    this  way,  we  find   a   table   as  follows: 


N 

0       1       1 

v) 

3 

4 

5 

6 

etc. 

b 

0 

5 

8 

9 

8 

5 

0 

etc. 

From  the  figui-e,  the  highest  value  of  b  is  seen  to  be  where  s  =  3, 
at  least  approximately.  Hence,  the  largest  box  will  result  by  taking 
the  side  approximately  3  ft.  The  shape  of  the  bottom  will  then 
be  a  square ;  and,  as  a  matter  of  fact,  this  is  the  shape  that  will 
give  the  greatest  volume  to  the  box. 


■■" 

"~ 

" 

1 

s 

" 

/ 

\ 

/ 

\ 

/ 

\ 

3 

j> 

'side 

It 

\ 

1 

\b  = 

=  s(6-J 

) 

\ 

, 

\ 

1 

1 

1 

1 

t 

-10 

1 

1 

1 

-L 

Fig.  i7. 


246    VARIATION  :    INDETERMINATE    EQUATIONS  [Ch.  IX 

EXERCISES    II:    CHAPTER   IX 

1.  Find    all    positive   integral    solutions    of    the    equation 

2x-\-7  y  =  50. 

2.  Find  some  integral  solutions  of  the  equation  3x—5y  =  7. 

3.  Find  all  integral  solutions  of  the  equation  -  4-^  =  3. 

3      4 

4.  Find  all  even  positive  integral  solutions  of  the  equation 
5  a-  +  7  2/  =  96. 

5.  Find  all  positive  integral  solutions  of  the  equation 
5  a;  +  7  2/  =  99,  such  that  x  is  odd  and  y  even. 

6.  Show  that  the  equation  5x+  7y  =  99  is  satisfied  by  all 
values  of  x  and  y  given  by  the  formulas  x  =  17  —  7  s,  y  =5  s-\-2. 

For  what  values  of  s  are  the  results  of  Ex.  5  obtained  ? 

7.  Show  that  the  equation  5x  —  6y  =  7  is  satisfied  by  the 
formulas  x=Gs  —  l,  y  —  5s  —  2.  Give  several  values  to  s,  and 
determine  values  of  x  and  y.     Also  solve  graphically. 

Find  the  least  positive  integral  solution  of  the  following 
equations : 

8.  7  x  —  Sy  =  17.  10.   3x—14:y  =  l. 

9.  12 x  +  5y  =  49.  XI.  49 x -53?/ =  17. 

12.  How  can  I  pay  a  debt  of  60  cents  in  quarters  and  dimes  ? 
Give  all  selections. 

13.  How  can  I  pay  $57  in  two-  and  five-dollar  bills  ?  Which 
method  uses  the  fewest  bills  ? 

14.  A  farmer  pays  a  debt  of  $4.50  by  giving  chickens  valued 
at  30  cents  each  and  turkeys  valued  at  80  cents  each  in  trade. 
How  many  of  each  sort  of  fowl  are  needed  ? 

15.  I  desire  to  weigh  100  pounds  by  means  of  8-  and  11-pound 
weights  placed  in  the  same  pan.     How  can  this  be  done  ? 


IJ.VIJG]  VAHIATIOX:    INDF/rKllMINATE  EQUATIONS     247 

16.  How  can  this  be  done  if  the  two  kinds  of  weights  are 
to  be  placed  in  opposite  pans,  the  3-pound  weights  being  in 
excess  ? 

17.  How  can  this  be  done,  the  11-pound  weights  being  in 
excess  ? 

18.  A  walks  n  hours  to  reach  a  neighboring  town ;  B,  start- 
ing from  the  same  place,  walks  5  hours  and  is  still  a  mile  from 
the  town.  If  each  man  walks  in  an  hour  an  integral  number 
of  miles,  how  fast  do  the  men  walk,  and  how  far  distant  is 
the  town  ? 

Note  that  one  condition  of  this  problem  is  the  limitation  of 
possible  speed  of  human  walking. 

19.  Find  a  number  that  leaves  a  remainder  of  11  on  division 
by  16,  and  of  4  on  division  by  11. 

Show  also  that  every  number  of  the  form  176  t  +  59,  where  t 
is  a  positive  integer,  has  this  property. 

20.  What  is  the  perimeter  y  oi  &  rectangle  having  one  side 
X  and  area  ^4  ?  Plot  the  relation  between  y  and  x  for  several 
values  of  A.  On  each  curve  note  the  value 
of  X,  which  makes  y  a  minimum.  What 
rectangle  of  given  area  has  the  minimum 
perimeter  ? 


21.  A  window  having  the  shape  of  a 
rectangle  surmounted  by  a  semicircle  has  a 
total  perimeter  of  200  inches.    What  dimen-   '  fig"^ 

sions  should  it  have  to  admit  the  greatest 
amount  of  light  ?     (Take  tt  =  y.)     What  is  then  the  area ? 

126.  Other  Cases.  Many  other  forms  of  indeterminate 
equations  arise.  Some  figures  have  been  drawn  on  pp.  30, 
182,  204,  to  illustrate  these,  particularly  in  connection 
with  quadratic  equations. 

In  general,  given  an  equation  containing  two  varying  quan- 
tities, u'e  give  several  values  to  one  of  them,  find  the  corre- 


248    VARIATION  :    INDETERMINATE    EQUATIONS  [Ch.  IX 


sponding  values  of  the  other^  and  plot  in  a  figure  the  points 
that  correspond  to  each  pair  of  values.  Finally  we  connect 
these  points  hy  a  smooth  curve. 

In  case  there  is  any  doubt  as  to  how  the  curve  should  be  drawn, 
we  simply  plot  more  points,  as  above,  until  the  curve  is  fairly  out- 
lined by  them. 

Among  interesting  curves  are  the  following : 
^x.  1.   y  =  kx^,  where  k  is  constant. 

This  arises  often  practically.  Thus, 
the  price  of  painting  a  square  surface 
varies  as  the  square  of  one  side.  p  =  k  •  s^ 
where  k  is  the  price  (in  dollars)  for 
painting  1  sq.  yd.,  and  where  p  means 
the  total  price  (in  dollars),  and  s  the 
length  of  one  side  in  yards. 

The  length  /  (in  feet)  of  a  pendulum 
varies  as  the  square  of  the  time  t  of 
vibration  (in  seconds): 

l  =  k.  t'^  where  k=  ^^^  =  3.26  (nearly). 

The  distance  d  (in  feet)  traversed  by 
a  body  falling  from  rest  varies  as  the 
square  of  the  time  t  (in  seconds)  it  falls : 

d=:k.fl  where  k  =  16.08. 

The  equations  of  the  form 
/  =  ax^  +  bx  -t-  c 

considered  in  the  previous  chapter  are 
of  the  same  type,  but  are  somewhat 
more  complicated  than  this  example. 

(a)  Let  us  draw  y  =  x-,  i.e.  the  case 
in  which  k  —  \. 


-J--             -r 

JU               ^ 

1                   1 

3:            o- 

± 

" '( & )     w,     r 

:-StSSiSI:: 

^fcpii-          or  H'^   1 

1                            If 

-4         ti 

a    IJ       _3n    Ll_t_    _ 

4    \-       -^    tt    f 

±j  ij        xti 

T           i  i 

V-               4^Z1 

I     _  ^...Kjt  i 

Ce)1    it          1i   Rc) 

M-^'    \i          //    kA^i 

~.\\         til 

lJJ    -i)t.i       -    _ 

AT      ui:    L 

-C        -^^ 

Ctt     ^-il 

ut    ~ji 

jit^lfC   - 

iX  M<ii 

Xa    ul 

is  ^tt 

k^Ja      , 

_SiL_--S 

0 

Fig.  49. 


Making  a  table  as  before,  we  find  : 


X     0 

1 

2 

3 

4 

5 

6 

7 

etc. 

-  1 

-  2 

-3 

-4 

etc. 

y    0 

1 

4 

9 

10 

1 

4 

etc. 

[Let  the  student  fill  in  the  blanks.] 


lL>(5]       VARIATION:   INDETERMINATE    EQUATIONS       249 

The  graph  is  as  drawn,  in  Fig.  40,  the  curve  marked  («). 
(A)   If  II  =  2  .r-,  we  find 


.,• 

0 

•1 

2 

;3      4 

5 

6 

etc. 

-  1 

-2 

-3 

-4  \    etc. 

y 

0 

2 

8 

18 

32 

2 

8 

The  graph  is  marked  (/>)  in  Fig.  49.  It  sliould  be  noticed  that  (h) 
{i.e.  the  curve  for  k='2)  is  just  twice  the  height  of  (a)  (i.e.  twice  the 
height  of  the  curve  for  t  =  1)  at  each  point. 

(c)  If  ^  =  ^  r-,  the  curve  is  just  half  as  high  at  each  point  as  is 
{(i).     [Let  student  make  a  table,  and  actually  draw  the  graph.] 

Ql)  [Let  the  student  draw  other  curves  to  represent  y —^x^, 
,j  =  4x2,  y  =  5a;2,  y  =  10.t2,  y  =  ^ x^  1/ =  ,'^ x\  y  =  -  x\  y  =  -  2x2, 
y  =  _ia;2.]  Different  colors  of  ink  or  pencil  may  be  used  to 
advantage  for  the  different  curves. 

All  these  should  be  drawn  on  the  same  sheet  of  paper  and  marked  so 
that  they  can  be  recognized. 

Ex.  2.    y  =  Jcx". 

This  equation  is  readily  plotted  as  above.  Only  the  figure  is  here 
given  ;  and  that  only  for  k  =  1,  k  =  2,  k  =  ^. 

[Let  the  student  make  a  table  for  each  of  these.] 


Fig.  50. 


250    VARIATION:   INDETERMINATE   EQUATIONS   [Ch.  IX 


Ex.3.    x^+jf  =  k\ 

(o)  If  k  =  1,  x2  +  rf  =  \. 

Take  a  point  P  and  mark  the  values  of  x  and  y  as  in  the  figures : 
X  =  OM,  y  =  MP. 

Now  OFf  +  MP^  =  UP^  since  OMP  is  a  right-angled  triangle. 
[The  student  should  know  that  the  square  of  the  diagonal  side  of  a 
right-angled  triangle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides.     See  Tables.] 

Hence,  x^  +  ?/2  =  op2.  b^t  ^.2  +  yi  ^  \  ;  hence,  OP'^  =  \ox  OP  =  \. 
Consequently  the  point  P  is  at  a  distance  1  from  0 ;  the  points  for 
which  this  is  true  lie  on  a  circle  of  radius  1  whose  center  is  at  0. 


Fig.  51. 

The  equation  x^  +  y^=l  is  represented  hy  a  circle  of 
unit  radius  about  the  origin  as  center. 

{h)  If  k  =  2,  the  reasoning  is  the  same  except  that  0P-2\  hence, 
3:-  +  y'^  =  4  IS  represented  hy  a  circle  of  radius  2  about  the  origin. 

(r)  In  any  case  x^  +  if  =  k''-  is  represented  hy  a  circle  of  radius  k 
ahuut  the  origin,  for  OP  =  k,  whatever  k  may  be. 

EXERCISES   III:    CHAPTER   IX 

Plot  the  following  curves : 

iOO  3    y=x'-7x  +  10. 


1-  y  = 


X- 


2.   y  =  x-\ 

X 


4.    2/  = 


50 


lU-x 


iL'ti]       VARIATION:    INDETERMINATE    P^QUATIONS       251 

50  ,       50  7.   y  =  x^  +  x. 

X       10  —  a; 

6.   y  =  x+\/x.  8.    ?/  =  2ar  — 9a;  +  8. 

For  various  values  of  k  draw  the  curves : 

9.   2/  =  -,-  11.    {x-ky  +  y'  =  l. 

X- 

10.    y  =  x-  +  2x-\-k.  12.    ar -h  (y  —  ^•)- =  1. 

For  various  values  of  k  and  ?,  draw  the  curves : 

13.  y^kx  +  ~-  15.    (x-ky+{y-lf  =  l. 

X 

14.  y  =  .r''  +  A-.r  + 1.  16.    ?/  =  ^'.*''  —  ^  V;c. 

17.  Represent  graphically  the  relation  between  the  ratios  of 
each  perpendicular  side  of  a  right  triangle  to  the  hypotenuse. 

Note.  These  ratios  are  called  the  "  sine  "  and  the  "  cosine  "  of 
one  of  the  angles  of  the  triangle  ;  it  is  suggested  that  they  be  denoted 
here  by  the  letters  s  and  c;  and  that  the  base,  altitude,  and  hypote- 
nuse be  denoted  by  b,  a,  and  h,  respectively.     Show  that  s  =  -,  c  =  -; 

h  h 

then  since  a^  +  6^  —  ^2^  show  that  s^  +  c'^  —  1,  by  dividing  both  sides 
by  h-\ 

18.  Represent  by  a  figure  the  relation  between  the  ratio  of 
the  hypotenuse  of  a  right  triangle  to  a  perpendicular  side,  and 
the  ratio  of  the  other  perpendicular  side  to  the  former. 

Note.  Denote  these  ratios  by  x  (  =  -  ]  and  by  n  =  -  )  (called 
"tangent  ").     Prove  first  that  x^  =  1  +  t'\ 

19.  Represent  by  a  figure  the  relation  between  the  ratio 
(c)  of  one  side  of  a  right  triangle  to  the  hypotenuse  and  the 
ratio  (x)  of  the  hypotenuse  to  that  same  side.  Show  first 
that  x  X  c=  1. 

20.  Find  the  maximum  rectangle  of  given  perimeter  k. 
(Solve  for  various  special  values  of  k ;  then  note  the  results, 
and  state  generally.) 


252       VARIATION:   INDETERMINATE   EQUATIONS 


SUMMARY   OF   CHAPTER   IX  :   VARIATION ;  INDETERMINATE 
EQUATIONS,  pp.  237-251 

Simple   Variation :  equivalence  to  proportion  between  variables ;  also 
to  equation  y  =  kx ;  figure,  straight  line.  §  122,  pp.  237-238, 

LinearVariation :   equivalence  to  equation  y  =  ax  -\-  b;  figure,  straight 
line. 

General  Linear  Equation:    figure  always  straight  line. 

§  123,  pp.  238-239. 

Inverse   Variation:   constancy  of   product   of  two   variables;    typical 
problem ;  Fig.  43,  inverse  variation  curve.     Exercises  I. 

§  124,  pp.  239-242. 

Indeterminate  Equations :   definition  ;  typical  problems ;   figures.     Ex- 
ercises 11  (for  graphical  solution).  §  125,  pp.  242-247. 

General  Indeterminate  Equations :    general  case  of  variation  ;  plotting 
curves  by  assignment  of  values  to  one  letter. 

Special  Types:    type  y  =  kx\  i.e.  variation   as  the   square;   y  =  kx'^; 
circle  x^  +  y'^  =  r^  center  at  origin,  radius  r.     Exercises  III. 

§  126,  pp.  247-251. 


CHAPTER   X.     SIMULTANEOUS    EQUATIONS 
INVOLVING   QUADRATICS 

PART    I.     ONE    LINEAR    AND    ONE    QUADRATIC 

127.  Introduction.  When  two  simultaneous  equations 
are  given,  one  of  which  is  linear,  the  other  quadratic,  it  is 
usually  best  to  use  the  method  of  substitution,  similar  to 
that  of  §  90,  p.  170. 

Ex.  1.  A  wagon  bed  is  to  be  made  to  hold  2  cubic  yards. 
It  must  be  4  feet  wide  and  six  times  as  long  as  it  is  high. 
Find  the  dimensions  of  the  wagon  bed. 

Let  X  be  the  height  of  the  bed 
in  feet  and  y  the  length  in  feet. 
Since  the  length  is  to  be  six 
times  the  height,  we  have 

(1)  y  =  ^x.  Fig.  52.  " 

The  volume  is  4  xy  (in  cubic  feet),  for  the  vohime  of  a  rectangular 
box  is  the  product  of  its  three  dimensions.  The  volume  is  to  be  2 
cu.  yd.  =  54  cu.  ft. 

(2)  .-.43;^  =.54,    or    2xy  =  27. 

(This  is  called  a  qnarlratic  equation  in  x  and  y,  for  the  sum  of  the  ex- 
ponents of  X  and  y  is  2.) 

Let  us  now  solve  the  equations  we  have  found : 

y^Qx,  (1) 

2  2-^  =  27.  (2) 

Substitute  6  x  for  y  from  (1)  in  (2)  : 

2  x(6  x)  =  27, 

or,  12x^=27,    whence,   x"^=:f,    or,    a;  =  ±|. 

253 


254 


SIMULTANEOUS   EQUATIONS 


[Cm.  X 


If  a:=  +  |, 

y=  Qx  —  Q, 

Hence,  the  result  is  x  =  |,  i/  =  9. 

Check : 

y  =  %x;  9  =  6x1  (correct). 
1xy  =  11\  2  •  1 .  9  =  27  (correct). 


If  a:  =  -|, 

?/  =  6  X  =  —  9. 
Hence,  the  result  is  x  =  —  |,  y  =  —  9. 

Check: 
y-%x\   -  9  =  6  ( -  I)  (correct). 

2x^  =  27; 
2  (  _  I)  (  _  9)  =  27  (correct) . 

In  this  example  only  the  first  set  of  answers  has  a  real  meaning. 
In  other  applications  of  these  equations  both  sets  of  answers  may  be 
useful. 

Check :  The  dimensions  of  the  wagon  bed  are  :  width  =  4  ft;  height 
=  li  ft;  length  =  9  ft.  The  volume  is  then  4  x  \\  x  9  cu.  ft.  =  54  cu.  ft. 
=  2  cu.  yd.,  as  required. 

F 


IT 

T 

± 

1: 

-/(I) 

t: 

r 

lt-L 

X\ 

I 

r 

I 

it 

jf 

t 

1 

1 

"^^         f2\ 

/ 

^s.  ^' 

\-  - 

*      "■    '-      -     L-,   ,_ 

'.-ZZZ-  X 

--it-         r^(X                                                             1 

To 

"^ ''  >-           3 

(2)     ^, ,          / 

■^       f 

\    if 

)    * 

~\ 

Cm 

Vf     - 

[A 

jf  1 

J I 

_l    T_ 

T    1 

/     1 

/       \ 

1    L 

i 

l\ 

±_     

Fig.  53. 

We  may  also  draw  a  figure,  Fig.  .53,  as  on  p.  2-39.  Thus,(l)  (.7  =  6  x) 
is  astraiglit  line,  as  drawn.  (Compare  pp.  141,  160.)  To  draw  (2) 
we  give  x  various  values  and  find  the  corresponding  values  of  y.  (See 
p.  239  and  §  126,  p.  247) 


127-1J8]      ONE   LINEAR   AND   ONE   QUADRATIC 


255 


X 

1 

2 

3 

4 

5 

6 

78 

9  10 

&c. 

-1 

-2 

-3 

-4&C. 

y 

¥ 

¥ 

¥ 

'i 

f^ 

fl 

-¥ 

11  (reduced) 

13.  T) 

G.To 

4.0 

8.37 

2.7 

2.41 

-13.5 

—  0.75 

[Let  the  student  till  in  the  blaulc  spaces  and  continue  the  table.] 


The  picture  for  (2)  drawn  from  tliis  table,  as  on  p.  254,  is  as  shown  ; 
it  has  two  parts  and  is  surely  not  straight.  In  fact,  comparing  with 
§  124,  p.  239,  we  see  that  (2)  is  a  curve  of  inverse  variation. 

Every  point  on  the  sfroigfit  line  (1)  f/ives  a  pair  of  numbers  that 
satisfy  y^Qx. 

Every  point  on  the  inverse  variation  curve  (2)  gives  a  pair  of  numbers 
that  satisfy  2  xy  =27. 

The  points  that  lie  on  both  (1)  and  (2)  satisfy  both  equations,  i.e. 
each  point  of  intersection  yives  a  pair  of  n  unifiers  that  are  a  pair  of 
simultaneous  solutions  «/(l)  and  (2).     Compare  §  86,  pp.  159-lGO. 

These  points  in  the  figure  are  P  and  Q- 

P  gives  (x  =  1^,  y  =  9)  Q  gives  (x  =  -  1^,  y  =  -  9),  which  are  in  fact 
the  answers  found  above. 

This  serves  as  a  check  on  the  work.  AYe  shall  later  find  this 
graphical  process  invaluable  in  solving  approximately,  in  difficult 
examples.     Compare  p.  276. 

128.  To  find  the  degree  of  a  term,  add  together  the 
exponents  of  each  of  the  important  {i.e.  unknown)  letters. 
(See  §  83,  p.  152.) 

The  degree  of  an  equation  is  the  degree  of  its  term  of 
higrhest  decree  after  it  is  freed  of  fractions  and  radical 
signs,  and  is  simplified  as  far  as  possible     (See  p.  152.) 

An  equation  of  the  first  degree  in  the  important  letters 
is  called  a  linear  or  simple  equation.     (See  p.  152.) 

An  equation  of  the  second  degree  in  the  important  let- 
ters is  called  a  quadratic  equation.  (See  p.  152  and  com- 
pare p.  208.) 

In  Part  I  of  this  chapter  we  consider  pairs  of  equations, 
one  of  whicli  is  linear.,  the  other  quadratic. 


256  SIMULTANEOUS   EQUATIONS  [Ch.  X 

129.  Rule.  To  solve  for  two  unknown  letters  in  a  pair 
of  equations,  one  of  ivldcJi  is  liyiear,  the  other  quadratic : 

(1)  Solve  the  linear  equations  for  y  {or  x^  in  terms  of 
X  (or  y). 

(2)  Substitute  in  the  quadratic  equation  for  y  (or  x')  the 
value  just  found. 

(3)  The  new  equation  will  be  a.  quadratic  in  x  (or  ?/),  or 
else  a  linear  equation  (in  rare  cases). 

(4)  Solve  this  equation  for  x  (or  y'),  and  substitute  both 
values  found  in  the  linear  equation  to  find  the  values  of  y 
(or  .r). 

(5)  Draw  the  figure  as  a  check  on  the  work. 

(6)  Substitute  each  pair  of  answers  in  the  original  equa- 
tions as  a  complete  check. 

Notice  that  there  will  usually  be  two  pairs  of  solutions,  for  we  solve 
a  quadratic  equation  during  the  process.  Care  should  be  taken  to 
pail-  off  the  values  of  x  and  (he  values  of  y  that  belong  together.  hi 
doing  this  the  figure  will  be  of  help  in  avoiding  errors. 

(2x  +  y  =  5,  (1) 

^''•^-  \x'  +  f  =  25.  (2) 

Solve  (1)  for  y:  y^5-2x. 

Substitute  (5  -  2  x)  for  y  in  (2)  : 

x2  +  (.5  -  2  xy  =  2.5, 
or,  a;2  + 25- 203;  + 4x2  =  2.5, 

whence,  5  x^  —  20  x  =  0,  or  5  x(x  —  4)  =  0. 

Hence,         5  x  =  0,  or  x  -  4  =  0,  that  is,  x  =  0,  or  x  =  4. 
If  X  =  0,  If  X  =  4, 

y  =  .5  [from  (1)].      8  +  7/  =  .5,  or  y  =  -  -^  [from  (1)]. 
The  pairs  of  solutions  are  therefore  (x  =  0,  ^  =  5)  and  (x=:4,  y  —  ~'d). 


129] 


ONE    LIXEAK    AND   ONE   QUADRATIC 


257 


Check: 
(  X  =  0  I 


2  x  +  y  =  5 

■f'  +  y-  =  ^■■> 

f  2  X  +  ^  =:  5 


2.0  +  5  =  5 
02  +  52  =  25 
2.4  +  (-3)  =  5 
[4-2 +  (-3)2=  25  J 


(correct). 


(correct). 


The  graph  is  as  shown  in  Fig.  54:  (1)  is  a  straight  line,  drawn 
as  in  §  80,  p.  140,  etc. 

(2)  is  a  circle  with  radius  5,  and  center  0,  as  shown;  see  p.  250, 
where  the  equation  x'^  +  y^  =  2o  is  studied.  The  values  (x  =  0,  ^  =  5) 
and  (x  =  4,  y  =  —  3)  as  we  have  paired  them  correspond  to  the  points  of 
intersection.  It  will  be  instructive  for  the  student  to  see  what  happens 
if  he  pairs  off  the  values  incorrectly :  thus,(x  =  0,  «/  =  —  3)  and  (x  =  4, 
</  =  5)  ;  do  these  give  points  on  the  curves?  Do  these  pairs  satisfy 
the  given  equations? 


V 

\ 

*v 

^ 

^ 

\ 

y 

^ 

' 

' — 

s 

(2 

/ 

\ 

/ 

*^ 

\ 

/ 

\ 

(1) 

^ 

0 

V 

\ 

\ 

\ 

\ 

^\ 

\ 

/ 

V 

.^ 

--• 

y 

V 

\ 

^ 

__ 

Fig.  54. 


Ex.  2. 

Solve  (1)  for  y  : 
Substitute  in  (2) 


lar^  +  w'  =  25. 


or, 
or. 


y 

?/  =  10  —  X. 
x2  +  (10  -  x)2  =  25, 
x2+  100-20x  +  x2  =  25, 
2  x2  -  20  X  +  75  =  0. 


(1) 
(2) 


258 


SIMULTANEOUS   EQUATIONS 


[Ch.  X 


The  solutions  of  this  equation  are  imaginary,  for  a  =  2,  J  =  —  20, 
c  =  75  in  the  notation  of  §  114,  p.  212. 

^2  -  4  ac  =  400  -  4  X  2  X  75  =  -  200. 
Hence,  there  is  no  value  of  x  among  numbers  we  know  at  present 
that  satisfies  the  equation.     (See  §  113,  p.  211 ;  also  Appendix,  §  31.) 

The  original  example  therefore  has  no  solutions.  This  is  clearly 
brought  out  by  the  graph  (Fig.  55) :  (2)  is  a  circle  of  radius  5  and 
center  O,  as  before;  (1)  is  a  straight  line,  as  shown.  It  is  clear  that 
these  two  curves  have  no  point  of  intersection  ;  hence,  there  is  no  pair 
of  numbers  that  satisfy  both  equations.     (See  p.  211.) 

Later  (see  Appendix,  §  37),  it  is  desirable  to  solve  the  quadratic. 
We  can  then  proceed  to  get  answers,  but  these  would  be  meaningless 
to  the  student  at  present. 


■ 

\ 

\ 

\ 

A 

\ 

\ 

(1) 

\ 

\ 

\ 

^ 

■% 

\ 

/ 

(2)\ 

\ 

\ 

\ 

0 

\ 

B 

\ 

y 

\ 

\ 

\ 

\ 

/ 

^ 

■-- 

X 

Ex.  X 

Solve  for  //  (1)  : 

Substitute  in  (2)  : 

or, 

Divide  by  25 : 


Fig.  .-)5. 

\x^  +  y^  =  25. 
25  -  4  a: 

..,(^^)^.25, 

25x2  _  200  X  +  625  =  225. 

x''^  -  8  X  +  25  =  9, 


(1) 
(2)    , 


k 


11'!.] 


OXK   LINEAR    AND   ONE   QUADRATIC 


259 


Transpose  9  ;  a:^  —  8  .r  +  16  =  0, 

or,  (x  -  4)-^  =  0. 

Hence,  x  =  4,  there  being  only  one  solution.     (See  §  112,  p.  209; 
tills  is  the  case  of  'UMjual  roots.") 

Since   x  =  4,    >j  =  o    from    (1).       The    only   pair   of    solutions    is 
therefore  (x  —  i,  ij  =  3). 

Cfieck: 


.       r  4  a:  +  .3  y  =  25  1      .  f  4  •  4  +  3  •  3  =  25  , 

'"    I  ,.?  ,    ../    ..^     \  S'^''^^   i  ,.,  ,   o.,      o.         r   (correct). 


r2  +  //•■;  =  25 


42  +  32  =  25 


s, 

s. 

\ 

^ 

1> 

s, 

\, 

^ 

-^ 

--> 

^^ 

/ 

\ 

J' 

(4 

.3) 

/ 

f 

s 

0 

\ 

/ 

\ 

\ 

\ 

\ 

s. 

\ 

/ 

s 

■x, 

— 1 

_^ 

y 

\ 

S, 

_^ 

_J 

Fig.  .")6. 

Figure  56  makes  clear  why  there  is  but  one  answer.  The  equation 
(2)  gives  the  same  circle  as  before,  i.e.  radius  5,  center  0.  The 
straight  line  (1)  goes  through  the  point  A  :  (x  =  0,  y  =  8|)  and  B : 
(x  =  6^,  ^  =  0).  It  also  goes  through  (a:  =  4,  y  =  3).  If  carefully 
drawn  as  shown  here,  it  is  seen  that  the  straight  line  and  the  circle 
have  only  one  point  P  in  common. 

The  straight  line  AB  is  said  to  be  tangent  to  the  circle. 

In  drmvinfi  a  graph,  if  the  equation  is  not  one  we  have 
already  studied^  it  must  he  plotted  out  by  points  as  in 
§  126,  pp.  247-248. 


260  SIMULTANEOUS   EQUATIONS  [Cii.  X 

EXERCISES   I:  CHAPTER   X 

Solve  the  following  pairs  of  equations  when  a  solution 
exists.  Note  in  each  case  whether  the  quadratic  in  one  vari- 
able obtained  in  the  course  of  the  solution  has  equal  or  distinct 
roots.  Always  draw  the  corresponding  graphs,  check  the 
numerical  results  by  substitution  in  the  given  equations,  and 
note  the  interpretation  of  two,  one,  or  no  roots  : 

*  1  I  ^'  +  /  =  34,  J  ^^    r  ar"  -  ir  =  8, 

*2  {•'«=^  +  /  =  25,  ^^-12  1^— r  =  i2, 

l;r  +  ?/  +  7  =  0.  \2x-y  =  &. 

*3.  . 


V 


-ar'-f  2/'  =  169,  ^3   f  9  .^-  + 16y- =  25, 

.  5  a;  + 12  ?/  =  169.  '  I  .^■  +  ?/  =  2. 

*4j^  +  2/==  =  17,  ^^\9x'  +  nf  =  2B, 

{x-\-y  =  '5.  \x-\-y  =Q. 

t  5.  I  ^y  ^  -^^'  15.  ^  ^  +  ^y=  1^' 


.T  -f  ?/  =  8.  1 2  a;  + .?/  =  8. 

'la;  +  ?/  =  6.  [a;-?/  =  -l. 

t  7.  I  ■'^'^^  ^  ^'^'  17.  ^  "^'^  +  ?^'  =  -  2) 


U-2/  =  2.  'U  +  2?/=l 

^3p.  +  3,  =  9,  [1  +  1=5, 

[a:2_7/2  =  8.  18.  j.^     y 

l4a;-3?/  =  l. 

■)■  9  f  •'^y  =  6,  19  f  •^-  +  y  +  -'W  =  11, 

'  \3x-2y=lQ.  '\x-y  =  l. 

§10.  {•^^  =  ^^'  20.  I  ^'-1^^^=!^' 
I  ?y  =  8.  U  -  3  .T  =  2. 

*  See  §  126,  p.  248.  J  See  §  131,  p.  269. 

t  See  §  124,  p.  240.  §  See  §  126,  p.  248. 


129-130]      ONE    IJNEAR   AND   ONE    Ql'ADKATIC  2G1 

130.  Ordinary  Quadratic.  The  following  special  ex- 
ample leads  to  a  new  ligure  for  graphical  solution  of 
quadratic  equations  in  one  letter. 

Ex.  1.     f  //  =  2  X-  +  8,  (1) 

\y^x^.  (2) 

Substitute  for  >/  fr;mi  (1)  in  (2): 

2  X  +  8  =  X-,     or,     a:2  -  2  X  -  8  =  0. 

Solving,  we  get  x  =  —  2,     or,     x  =  +  4. 

If  ar  =  -  2,  If  x  =  +  4, 

2/  =  2  x  +  8  =  2  .  (-  2)  +  8  =  4.  y  =  2  •  4  +  8  =  16. 

Check' : 

x  =  4    ]        (y  =  2x  +  S]  ri6  =  2.4  +  8] 

MiM  t  give    -i  \  (correct). 

2^  =  16  I        \y=x-^  J*  [16=4-^  j^  ^ 

The  figure  is  as  shown  on  p.  264  :  (1)  is  a  straight  line  (draw  it) ; 
I    (2)  is  as  shown  (see  §  95,  p.  183). 

From  the  figure  there  are  two  points  of  intersection,  P  :  (x  =  —  2, 
y  =  4)  and  Q  :  {x  —  4,  y  —  16)  ;  these  pairs  of  numbers  are  therefore 
I    solutions,  as  we  found  above. 

From  this  example  it  is  clear  that  the  solution  of  the  quadratic 

equation 
^  x2-2x-8  =  0 

gives  the  same  values  of  x  as  are  given  for  x  by  the  simultaneous 

!  [  ^  =  2  X  +  8. 

.     Likewise,  the  values  of  x  found  in  x^+px  +  q  —  O  are 
also  found  from  r  .> 

y  =  -px-q, 

for,   substituting    for  y,  the  last  pair  give    the    previous 
equation.     The  advantage  of  this  graphical  picture  over 


262 


SIMULTANEOUS   EQUATIONS 


[Cii.  X 


those  in  Chapter  VIII  is  that  the  curve  y  =  ^  is  the  same 
for  all  quadratics  by  this  method.  This  curve  being 
drawn  once  for  all,  nothing  remains  but  to  draw  the 
straight  line  y  =  —  px  —  q. 


T^             T 

t                4     + 

V-                    4      ^ 

\            i  j- 

"t           iJ^ 

\        ^  ti 

4                t 

/ 

'               ol 

,                       / 

k            t\ 

t        ji 

X        tt 

\      -it 

A       t  t 

\      2    ^ 

-L    ^  i 

\      ^B    A 

\  t     1 

\j-      t 

py        1 

A        1 

t\     2   ± 

-/it 

^'       ^z 

7          0 

T 

J 

t 

J. 

X 


Fig.  57. 

Several  examples  may  be  drawn  in  the  same  figure,  as 
follows : 

Solve  the  following  list  of  examples  graphically  : 
{^)2^-^x^\={).  Seep.  203.   (4)  x--4.r+5=0. 
(2)     ar^-4a;+l  =  0.  See  p.  207.   (5)  ar'+2  =0.     See  p.  211. 

(8)     ar-4x-+4  =  0.  Seep.  209.    (6)  x--  +  2a-+5  =  0.     Seep.  211, 


13U] 


ONE    LINEAR    AND   ONE    QUADRATIC 


263 


In   (1)   it  is  desirable  first  to  reduce  the  coefficient  of  x^  to  unity 
by  dividing-  both  sides  by  2 ;  this  gives 

(1)      x2-|x  +  2  =  0. 
Y 


1 

(2) 

/ 

1 

1 

^v 

1 

1 1 

{■i 
1 

'  1 

/ 

1 

P 

) 

/ 

j  . 

1 

ji 

/ 

II 

II 

1 

II 

-Ij^ 

1 1 

1 

T 

» 

ii 

1= 

11 

1, 

II 

• 

l\ 

V 

J) 

1 

i^ 

\ 

\ 

i 

^ 

\ 

li 

1 

\ 

\ 

r/ 

1 

\ 

s 

1 

1 

V 

1 

- 

\ 

\ 

v 

\ 

\ 

1 

1 1 

V 

\ 

jUI 

\ 

^ 

1 

-i 

0 

- 

V 

\ 

y 

? 

i 

0 

\ 

/ 

5) 

\^ 

(5) 

^ 

i) 

\ 

\ 

rf 

\ 

\ 

0) 

( 

n 

m 

Fig.  58. 
As    above,    (1)  corresponds    to    solving    tlie    simultaneous    pair 

_   ,.2 


I 
II 


264  SIMULTANEOUS   EQUATIONS  [Ca.  X 

I  is  the  figure  drawn  above  ;  II  is  a  straight  line,  marked  (1')  in 
the  figure  (draw  it).  Tlie  points  where  these  meet  are  the  simul- 
taneous solutions  of  I  and  II;  the  values  of  x  at  the  points  of  meeting 
are  the  solutions  of  (1),  namely  x  =  h,  and  x  =  4  (see  p.  203). 

The  other  examples  are  solved  graphically  by  drawing  in  the  figure 
the  lines 

(2')     y  =  ^x-\,  (4')     ^  =  4x-5, 

(;3')     3/  =  4a._4,  (5')     y  =  -2,  (G')     2/  =  -2x-r3. 

The  line  (2')  meets  the  curve  at  two  points  ahou(  x  =  \  and  x  =  3.7. 
This  solution  is  not  precisely  correct,  but  is  approximately  so.  (See 
p.  207.) 

The  line  (3')  meets  the  curve  in  only  one  point ;  then  (3)  has  only 
one  solution,  x  =  2.     (See  p.  209.) 

The  line  (4')  does  not  meet  the  curve  at  all ;  hence  (4)  has  no 
solution.     (Imaginary  case ;  see  p.  211.) 

The  other  examples  should  be  compared  with  their  previous 
solutions. 

EXERCISES   II:  CHAPTER   X 

Draw  the  graphs  for  all  the  following  examples  on  one  dia- 
gram; classify  them  according  to  the  number  of  answers; 
estimate  the  solutions,  if  there  are  any,  from  the  figure ;  check 
by  solving  the  equations,  or  if  there  are  no  solutions,  by  com- 
puting ii"  — 4  ac: 

7.  3a;2-8a;  +  5  =  0. 

8.  1  a?  —  x+l=0. 

9.  G.r-'+6.x-+|  =  0. 

10.  .r  —  .T  -f  1  =  0. 

11.  X-  +  X  —  2  —  0. 

12.  a-2-r,.r  +  l=0. 

13.  Draw  a  figure  for  solving  the  equation  a^— 4a;+g  =  0 
for  various  values  of  q  from  g  =  0  to  q  =  10  by  the  method  of 
§  130.  Classify  the  character  of  the  solutions  of  the  equation 
(§  115)  according  to  the  value  of  q,  estimate  solutions,  and  check. 


1. 

x'-<dx  +  2Q  =  0. 

2. 

x-  -  3  a;  -f  5  =  0. 

3. 

ar'_3a^_5  =  0. 

4. 

ar'  +  8.T-fl5  =  0. 

5. 

X-  -f  8  a;  — 15  =  0. 

6. 

4.^^  +  9_12a•  =  0 

130]  ONE  linp:ar  and  one  quadratic         .265 

REVIEW   EXERCISES    III  :     CHAPTER  X 

Solve  the  following  pairs  of  equations : 


x+y 


—  n 


'  [32^-5  q  =  l.  (2zy-z-  y=  4, 


_    f     m-  —  7nn  =  3, 
3.  i  ' 


7.         ^  •' 

\3z-10y=5. 


l-»i-n  =  4.  Q(x^  +  f  +  x-\-y=5i, 

(2v--3n-  =  3S,  '\x-y  =  4:. 

4.  ^  ' 


[2v-3u  =  4. 


n 


5   lk  +  l  +  M  =  5, 
'\5k-2l==l.  [3n-r  =  6. 

10.  The  diagonal  of  a  rectangle  is  13  inches  long.  What 
are  its  dimensions,  if  one  side  is  7  inches  longer  than  the 
other  ? 

11.  It  takes  2  hours  longer  for  one  pipe  to  empty  a  tank 
than  for  a  second  pipe  to  empty  an  equal  tank.  Both  pipes 
together  can  fill  either  tank  in  1  hour  20  minutes.  How  long 
would  it  take  each  separately  to  do  so  ? 

12.  The  fence  around  the  outside  of  a  walk  5  feet  wide  sur- 
rounding a  park  lot  is  340  feet  long ;  the  area  of  the  lot  itself 
is  5000  square  feet.     What  are  the  dimensions  of  the  lot  ? 

13.  AVhat  integer  can  be  taken  such  that  the  sum  of  all 
integers  from  10  lip  to  and  including  the  chosen  one  shall  be 
1230? 

Solution.     Let  n  be  the  chosen  number.     Then  the  sum  is 
10  +  11  +  12  +  ...  +  (»i  -  1)  +  n  =  1230. 

Simply  reversing  this,  we  get 

„  +  („  _  1)  +  („  _  2)  -  +  11  +10  =  1230. 
Adding,  (10  +  ri)  + (10  -f  n)  +  -  +(10  +  n)  +  (10  +  n)  =  2460. 


266  SIMULTANEOUS   EQUATIONS 

If  there  are  t  terms,  we  have 

t  (10  +  n)  =  2160. 

But  it  is  easy  to  see  that 

10  +  t -1  =  11. 

Solving  these  two  equations,  as  above,  we  find, 
n  =  50,  t  =  41. 

14.  If  we  add  together  the  numbers  obtained  on  starting 
with  —  I  and  increasing  by  unity  successively,  where  must  we 
stop  in  order  to  have  the  sum  8  ? 

15.  Solve  Ex.  9  of  Chapter  VIII,  Exs.  VI,  p.  216,  by  the 
use  of  two  unknowns. 

16.  Solve  Ex.  18  of  Chapter  VIII,  Exs.  VI,  p.  217,  by  the 
use  of  two  unknowns. 

17.  Solve  Ex.  30  of  Chapter  VIII,  Exs.  VI,  p.  219,  by  the 
use  of  two  unknowns. 

18.  Solve  Ex.  54  of  Chapter  VIII,  Exs.  VI,  p.  221,  by  the 
use  of  two  unknowns. 

19.  Solve  Ex.  51  of  CUiapter  VIII,  Exs.  X,  p.  233,  by  the 
use  of  two  unknowns. 

20.  Solve  Ex.  56  of  Chapter  VIII,  Exs.  X,  p.  233,  by  the  use 
of  two  unknowns. 

As  suggested  in  Exs.  15-20,  many  problems  solved  in  pre- 
vious chapters  by  means  of  one  unknown  may  now  be  solved, 
in  some  cases  more  expeditiously,  by  the  use  of  two  or  more 
unknown  numbers ;  it  is  recommended  that  many  problems 
from  previous  lists  be  now  solved  in  this  manner  under  the 
guidance  of  the  teacher. 


PART   II.     SIMULTANEOUS   QUADRATICS 

131.  Simultaneous  Quadratics.  Two  quadratic  equations 
which  are  both  to  hokl  true  are  called  a  pair  of  simultane- 
ous quadratics. 

We  can  often  solve  such  pairs  by  methods  similar  to 
those  of  §  l-2l»,  p.  256,  and  §§  86-89,  pp.  159-172.  The 
following  examples  will  illustrate  these  methods : 

T,     1     n-  ra-  +  r=10,  (1) 

Solve  (2)  for  x^:  x"^  =  Q  y. 

Substitute  G  y  for  x^  from  (2)  in  (1)  : 
(3)  3/2  +  6^  =  16. 

Complete  the  square  :  y-  +  6  ^  +  9  =  25, 
or,  ?/  +  3  =  ±  5. 

Hence,  either,  y  =  2,  or  y  =  —  8. 

If  ^  =  2,  It>=-8, 

x2  =  12,  x2  =  -  48, 


and  X  =  ±  Vl2.  x  =  ±  V—  48  (imaginary). 

The  real  solutions  are  (x  =  +  VT2,  y  =  2)  and  (x  =  —  VrJ,  y  =  2). 

The  other  expressed  solutions  are  meaningless  to  the  student  at 
present.  They  are  (x=  V—  48,  y  =  —  8)  and  (x  =  —  V—  48,  y  =  —  8); 
we  shall  not  regard  them  as  solutions  here  (see  Appendix,  §§  31-38). 

Check  for  real  sulutions: 

fx  =  ±vl2l.      (x^  +  y'  =  l6\     .        [12  +  4  =  16] 

\         "T  r  11^  -^  ,  \  gives  {  \    (correct). 

Graph.  The  figure  is  easy  to  draw,  for  each  curve  was  drawn 
above  :  (1)  is  a  circle  of  radius  4  about  0  as  center  (see  p.  250)  ;  (2) 
is  a  curve  of  the  kind  drawn  in  §  126,  p.  248. 

267 


268  SIMULTANEOUS   EQUATIONS 

The  table  for  y  =\  x-  is 


[Ch.X 


X     .    . 

0 

±1 

±2 

±3 

±4 
2f 

±5 
H 

±6 
6 

±7 
8^ 

±8 
lOf 

±9 

±10 

etc. 

±15 

etc. 
etc. 

±20 

y  •  ' 

0 

1 

6 

m 

16| 

etc. 

371 

66f 

Y 

\ 

1 

\ 

:      ±      t 

1 

-  .in                            -1 

V               7 

L 

V 

1 

\ 

T 

N. 

-•^            -/- 

\ 

-T,                              f 

\, 

Q^ 

-~^^N^ 

's 

/ 

^ 

.» 

/ 

-         ^ 

<rff 

? 

'<,  ^ 

0     ,^'      '•     \ 

\ 

i 

\ 

/ 

:^^^ 

X 


Fig.  59. 

The  whole  figure  is  as  drawn.  From  it  we  note  the  two  points  of 
intersection,  P  and  Q:  P  is  (x  =  3.5  (about),  y  =  2)  and  Q  is 
(x  =  —  3.5  (about),  z/  =  2) . 

These  agree  with  the  results  above  as  closely  as  we  could  expect; 
in  fact  Vl2  =  3.464  •••  as  will  be  found  by  the  ordinary  process  for 
square  root  in  arithmetic. 

The  solution  of  this  example  illustrates  the  method  of  suhslilution- 
(Compare  p.  170  and  p.  256.) 


Ex.  2. 


x"  -  y-  =  7. 


(1) 
(2) 


We  proceed  either  as  before,  by  substitution,  or  as  follows : 


■1] 


SIMULTANEOUS   QUADRATICS 


269 


1 
1 

xr  ^-  y-  =  25 

x:-  -  11-  =    7 

2 

x2  =32 
a;2          =16 

j;            =±4 

1 

a:2  +  ^2  =  25 

-1 

x2-y2=    7 

2  ^2  ^  18 

.y2=    9 

2/  =±3 

The  possible  combinations  are  (compare  example  1,  pp.  256-257): 

ra:  =  +  4  rx  =  4-4  (x  =  -4  (x  =  -i 

A:   \  ;      B:   \  ;     C:  ;     D : 


y--3 


The  figure  shows  these  points  clearly:  equation  (l)is  the  circle 
drawn  before  (radius  5,  center  0)  ;  equation  (2)  gives  the  following 
table : 


y 

±0 

±1 

±2 

±3 

±4 

±5 

±6 

±7 

±8 

±9 

&c. 

X 

±V7 

±V8 

±vTr 

±Vl6 

±V23 

±V'32 

— 

X  (reduced) 

±2.66 

±2.8 

±3.3 

±4 

±4.8 

±5.7 

and  the  figure  is  as 
shown  in  [(2)  in  figure]. 
The  points  A,  B,  C, 
D  of  intersection  corre- 
spond to  the  pairs  of 
solutions  just  found. 


Check  for 


rx  =  ±4i 

r  42  +  82  =  2.5, 

1 42  _  32  =  7. 

The  solution  of  this 
illustrates  the  method 
of  addition  or  siihtrnction. 
(Compare  p.  163.) 


- 

- 

- 

\ 

( 

) 

V 

(2)\ 

A 

N 

n-i) 

\ 

/, 

1^ 

/ 

^^ 

/ 

/ 

' 

\ 

0 

f 

1 

1 

- 

y 

' 

'\ 

V 

/p 

\\ 

/ 

K 

' 

V 

/ 

' 

■v 

"- 

f^ 

^ 

\ 

/ 

\ 

Fig.  60. 


It  is  always  successful  for  pairs  of  equations  of  the  type 


ax-  +  h]f-  =  -c. 


270  SIMULTANEOUS   EQUATIONS  [Cii.  X 

In  solving  exercises,  ingenuity  is  sometimes  required. 
Often,  however,  some  simple  process,  usuall}^  the  method 
of  substitution  or  the  method  of  addition  or  subtraction  will 
suffice  in  the  following  exercises.  (See  also  Chapter  XII, 
p.  318,  and  Appendix,  §§  39-42.) 

EXERCISES    IV  :   CHAPTER   X 

Solve  the  following  equations;  always  draw  the  figure: 


I  ic^  +  xy  =  56, 

2  X-  -  xy  =  35.  I  ^^2  ^  2iiq  =  21, 

[SuoGKSTiON.      Solve  first  for  [  2  jxj -}- q- =  16. 

'  ■  -'  [Suggestion.       Multiply     the 

C  o  .2 K   ^ ir,  first  equation  by  16,  the  second  by 

4a<  =  161. 


4.     -1  21,  and  subtract ;  factor  the  result- 

15^-      '     ■      ' "' 


ing  equation. 

Another  method  is  to  set  q  = 
\u-\-V=  oU,  ^^ .  solve  both  equations  for  p'^  and 

t  uv  =  7.  compare  results.] 

[Suggestion.      Multiply     the       -^^      |  •'»•'  + -^7/ +  .'/'  =  ^  ^> 

second  equation    by  2,  and  com-  I  .ir  —  //- =  5. 

bine  by  addition  and  subtraction  (-.210      .102      c 

with  the  first  equation.)  12.     { 

I  2  a;2  +  3  .x-!/  +  4  ?/2  =  9. 

g      |m  +71- =  29,  [Suggestion.      Eliminate   the 

I  mn  =  10.  terras  in  xy."] 

132.  Raising  the  Graph  Vertically.  It  will  often  hap- 
pen that  there  will  be  no  solution.  Examples  of  tlie 
various  possibilities  follow  in  §  133. 


l:n-lo2] 


SIMULTANEOUS   QUADRATICS 


271 


Let  us  first  notice  the  effect  of  writing 

(1)  y=6->:'-+k 
in  place  of 

(2)  Z/=B-^' 

in  example  1,  §  lol,  where  k  is  some  fixed  number.     If  we  write 

(•0  1/  =  i  ^^  +  2, 

for  example,  the  curve  in  the  figure  will  be  just  2  units  vertically 
above  the  curve  for  equation  (2),  since  each  value  of  x  makes  y  just 
2  units  greater. 

The  figure  is  drawn  for  the  equations  (2),  (3),  and  also  for 


(4) 


X-  +  5, 


(5)     Z/  =  i  ^^ 


There  is  a  curious  optical  illusion  about  such  figures,  which  may 
deceive  the  student;  although  they  may  not  seem  to  be  so,  these 
curves  are  the  same  shape  and  size,  and  any  one  is  formed  from  any 
other  by  simply  raising  it  or  lowering  it.  The  student  will  convince 
himself  of  this  by  cutting  out  a  piece  of  paper  to  fit  in  one  of  them 
and  then  raising  or  lowering  it. 


Fig.  61. 


272  SIMULTANEOUS   EQUATIONS  [Ch.  X 


133. 

Various  Possibilities  Illustrated.     Consider  now 

the 

Uowing  exuniples : 

I. 

'  x--{-ir  =  16, 

(1) 
(2) 

y          (x'  +  ?f=16, 

(1) 

(2) 

II. 

(1) 
(2) 

yj      [-  +  Z/^  =  16, 

(1) 
(2) 

III. 

|•^■"  +  r  =  16, 

(1) 

(2) 

vii.   \^^^r  =  ie, 

(1) 
(2) 

IV. 

|ar^  +  2/'  =  16, 

(1) 
(2) 

VIII.  {-  +  ^^^=16, 

\y=l-x^-10. 

(1) 
(2) 

These  eight  examples  are  all  very  similar  to  example  1,  p.  267  ;  and 
they  may  be  solved  in  the  same  way.  But  a  glance  at  the  figure 
shows  that  Exs.  I  and  II  have  two  solutions  each  (A  and  B,  and  C 
and  D,  in  Fig.  62)  ;  Ex.  Ill,  only  one  solution  (E  in  Fig.  62)  ;  Exs. 
IV'  and  V,  no  solutions;  Ex.  VI,  three  solutions  (F,  G,  H,  in  Fig.  62)  ; 
Exs.  VII  and  VIII,  no  solutions.  For  the  equation  (1)  is  the  same 
in  all  of  them,  while  equation  (2)  differs;  in  the  figure,  equation 
(1)  is  a  circle  of  radius  (4)  about  0,  while  equation  (2)  is  the  curve 
of  §  181  moved  upward  or  downward.  The  student  will  see  just 
when  there  are  solutions  by  cutting  out  a  i)iece  of  pajier  the  shape 
of  the  curve  and  moving  it  up  or  down. 

Let  us  solve  Ex.  VI  by  the  algebraic  methods  to  check  these 
results: 

In  Ex.  VI  put  x2  =  6  ?/  +  24  from  (2)  in  (1)  ;  then 
?/2  +  6?/ +  24  =  16. 
Solving  this  equation,  we  find  ;/  =  — 4or^  =  —  2. 
If  y  =  -  4,  Uy=-2, 

_  4  =  1  x2  -  4  from  (2),         _  2  =  ^  x^  -  4, 
or  j:^  =  0,  or  X  =  0.  or  x^  =  12,  or  .r  =  ±  Vl2. 

^,       ,  f  X  =       0  f  .r  =  +  VT2  f  X  =  -  Vl2 

The  throe  answers  are    \  ;       ■,'  ;       <  ; 

b--4        \!/=-2  \y=-2 

which  are  the  points  marked  F,  G,  IT,  respectively,  in  Fig.  62.  Each 
of  the  other  examples  may  be  solved  in  a  similar  manner;  the  results 
will  be  found  to  agree  with  the  graph. 


SLMULTAXEOUS    QL'ADllATICS 

r 


273 


— 

— 

I 

1 

1 

1 

1 

! 

' 

(V, ) 

\ 

V 

i     1  L 

1 

ax 

lt/^ 

'i 

t 

""'i 

\ 

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^ 

i  \ 

/ 

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nanif 

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/ 

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(U. 

\\\  \ 

^ 

y 

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a.) 

w  \ 

\ 

w  \ 

\ 

/ 

tllHx^  1 

' 

\W 

\ 

/     1 

:  J 

\  w  \ 

\ 

/      / 

// //  /  1 

\\  WW 

\ 

ax.>/     / 

////  ''^''" 

.) 

W  WW 

V 

\ 

1 

- 

//, 

www 

\ 

\ 

/ 

// 

nViWW 

1 

V 

f^   /  / 

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\  \  i\\  1  w  ^ 

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\  \ 

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ll  11  {.V 

11 . 

) 

\  \  w  \\ 

^.  \ 

^ 

' 

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II   1 

-^ 

1  \  \  \\  v 

\  \ 

s 

/ 

(V'.)/l   /// 

III 

\\\  \\  y 

V 

y 

/' 

1  /    //I 

y/  /'-'^■' 

\  \  \  « 

\ 

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y  1 

/    //  / 1 

I    A  / 1 

\\  \  \  W 

N. 

i^v. 

y 

1 

/  '  //  /  / 

1  11' 

\\  \   \       H 

i\ 

1 

f\  nil 

1    I  ]  \ 

\\  \   \       H 

.  1  s. 

/ 

'    '         f   i 

1     /  /(XI.) 

1 

\  \    \  M 

\    > 

k 

>( 

IV Oy/ //    ti 

\   1 

\\\i  \ 

!^?U^ 

\. 

, 

/ 

11   1 J    II 

\\\   \ 

r\\%^ 

^^•^^ 

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h 

If    11   if 

III 

vw  \ 

W  \\  \ 

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1  /  \  1  1  \   II 

I    f  /( XII  ) 

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w  \ 

t^ 

n^ 

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1 

\\\\ 

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// 

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11 

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k\N 

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rli 

Vn 

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cviii.)/ 

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fir 

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^^ 

\l\l 

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^^^■1 

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t 

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ki.) 

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■^ 

■/(XM 

) 

X. 

_ 

Fig.  62. 


274  SlAUi/rANEOL'S    KQUATJONS  [Ch.  X 

lu  the  same  figure  we  have  drawn  the  new  circle 

x^  +  y^=  100, 

which  is  a  circle  of  radius  10  about  O.  This  shows  some  of  the 
details  lost  because  of  small  size  in  the  other  figure.  The  figure 
using  the  larger  circle  shows  graphically  the  solutions  for  the  follow- 
ing examples : 

x^-  +  /  =  100,  VTT'      I  ^'  +  r  =  100, 


1'.  -^    -r-^    -^^'^,  yjj 


1  o^ 


Ij,      fa;-  +  r  =  100,  yjjp      l.r-f/=100, 

IIP.     f^'  +  /=100,  j^,_     |:t'^  +  r  =  100, 

\y=lx'  +  5.  '     \y=^x'-15. 

y,      |.t2  +  i/-  =  100,  -j^j,      fa;2  +  r'  =  100, 

Yi,    [x2+r  =  ioo,  XII '.    f^^''  +  r  =  ioo. 

These  are  seen  by  the  figure  to  have  the  following  number  of  solu- 
tions (which  can  be  judged  approximately  from  the  figure):  1',  two  (at 
A'  and  B')  ;  11',  two  (where  in  Fig.  ?)  ;  III',  two  (where  in  Fig.  ?) ;  1 V, 
two ;  V,  one  (at  /) ;  IX',  none ;  VI',  two  at  G'  and  H' ;  Vll',  tivo  ;  VIII', 
three,  at  L,  M,  N;  X',four,  at  P,  Q,  R,  S ;  XI',  two,  at  T,  V;  XII',  7,one. 

Of  these  the  most  remarkable  are  X'  with  four  solutions,  and  XI' 
with  its  curious  two  solutions.  There  are  similar  ones  with  the  small 
circle ;  but  they  cannot  be  seen  readily  in  the  figure  because  of  the 
small  size.  The  position  in  XI'  is  determined  as  follows,  so  that  there 
shall  be  just  two  roots  at  T  and  U.     We  try  y  =  I  x^  +  k  and  solve : 

(1)  r  .t2  +  ^/2  =  100, 

(2)  \y  =  lx'^  +  k. 

From  (•_>),  x^=Gy-Qk. 

Substitute  in  (1)  :  y"^  +  G  y  -  G  k  =  100, 

or,  ?/2  +  6  //  +  (  -  100  -  «  k)  =  0. 


i:?;5]  SIMULTANEOUS   QUADRATICS  275 

The  condition  tliat  this  equation  should  have  equal  roots  is  that 
li-  —  i  ac  —  0 ;  we  have  a  =  1,  h  —  6,  c  =  —  100  —  6  ^•;  hence, 
//2  -  4  ac  =  36  -  4  (  -  100  -  6  k)  =  0,  or  yl-  =  -  18J, 
which  was  the  vahie  taken  in  XT. 

It  is  to  be  noticed  that  the  vahie  of  y  may  be  real  although  x  is 
imaginary,  and  vice  versa,  and  as  in  the  first  case  solved  (p.  267). 

The  student  will  find  other  possibilities  by  cutting  out  the  figures 
shown  above  and  moving  them  around  across  each  other.  Some  of  these 
possible  positions  correspond  to  fairly  complicated  pairs  of  equations. 

Other  curves  may  be  tried,  as  suggested  in  the  exercises  below. 

A  complete  study  of  the  possibilities  —  at  least  a  full  understand- 
ing of  them — cannot  be  hoped  for  here.  It  is  only  after  a  study  of 
Analytic  Geometry,  in  which  such  questions  are  discussed  at  length, 
that  the  student  will  really  appreciate  all  that  is  involved. 

A  few  special  rukii  are  sometimes  given ;  these  are  really  not 
particularly  useful.     See  Appendix,  §§  39-42. 

EXERCISES   V:    CHAPTER   X 

Draw  figures  for  Exs.  1-5,  note  the  number  of  solutions,  esti- 
mate their  numerical  values,  and  solve  algebraically: 

■   l?/2  =  2x-16.  ■  \y-  =  2x-^. 

(Xo  real  solution.)  (One  set  of  solutions.) 

r  .ir'  -  V-  =  16,  r  x^  -  v' = 16,  f  y?  -i/=l  6, 

3.   -j  •  4.   ■(  ^  5.        ,      ^ 

(How  many  sets  of  solutions  in  each  case?) 


Similarly  the  following  exercises  (6-8)  : 

f  .r-  —  if-  =16,  {  x^  —  y-  =  16,  {  or  —  if=\  (5, 

6.   ^  7.  ^  8.   ■!     ,      ^' 

I  .^•2  -f  if  =  9.  Vx-  +  y-  =  l(j.  I  ar  -f  ?/'  =  25. 

Similarly  the  following  exercises  (9-16)  : 

I  x^  +  /  =:  49.  I  {x  -  5)2  -f  r  =  49 


276 


SIMULTANEOUS   EQUATIONS 


[Ch.  X 


12. 


13. 


x^  —  y^  =  l, 
(.^■-7)-+7/-  =  49. 

x"  -  ir  =  1, 

(x  -  8)-  4-  r  =  49. 


14. 


15. 


16. 


x^-if  =  1, 

(a;  _  10)2  +  2/2  =  49. 

o?  —  i/=  1, 

(x-- 15)2 +  7/2  =  49. 

^2  —  ?/2  =  1, 

(x- 20)2 +  2/2=49. 


Treat  the  following  similarly,  except  that  an  algebraic  solu- 
tion need  not  be  obtained ;  instead,  check  the  estimated  results 
by  substituting  in  the  equations : 


17. 


xy  =  6, 
?/  =  3  x\ 


18. 


xy  =  6, 

y  =  3  .t2  —  9. 


19. 


2/  =  3  cc2  - 18. 


134.  Graphical  Solution.  Graphical  methods  of  solution 
have  been  mentioned  above  in  practically  all  cases  as  a  con- 
venient check  and  as  a  method  for  finding  approximate 
answers.  There  are  many  problems  so  difficult  that  no 
method  except  the  graphical  one  is  really  convenient,  or 
even  possible,  with  the  student's  present  knowledge.  Such 
is  the  example  below. 

x-^y  =  l,  (1) 

x-\-f  =  ll.  (2) 


Ex.1. 


This  pair  of  equations  defies  all  the  methods  of  elementary  algebra 
except  the  graphical  method. 

Equation  (1)  may  be  written  in  the  form 

(1)  y  =  _  x2  +  7. 

A  table  of  values  of  x  and  y  is : 


X 

±0 

±1 

±-i 

±:\ 

±4 

i^) 

±6 

±7 

±8 

±9 

±10 

etc. 

y 

7 

fi 

:?       -  2    -  0 

-18 

l:];3-l;31] 


SIMl'LTAXEOrS   QUAl^I^A'lICS 


277 


and  the  figure  is  as  shown  in  Fig.  Co.  It  is  the  same  as  the  curve 
of  Fig.  31,  p.  18:5,  turned  upside  down  aud  then  raised  vertically 
7  points.     Equation  (*J)  may  be  written 

X-  -  y-  +  11. 
A  table  of  values  of  x  aud  ^  is : 


£ 

11 

±  0 

10 

±  1 

7 

•2      -  5    -  U 

1 

±2 

±3 

±4 

±5 

±G 

±7 

±8 

etc. 

aud  the  figure  is  as  shown.  It  is  the  same  as  tlie  curve  of  Fig.  31, 
p.  183,  turned  on  its  side  (/.e.  turned  through  UU^)  and  then  moved 
horizontally  11  points  to  the  right. 


r 


— 

— 

— 

. 

- 

/ 

N 

/I 

Vf^ 

1 

^1'^ 

\ 

--f4- 

\a 

/ 

\ 

■— 

..... 

(2 

/ 

\ 

— 

-> 

/ 

0 

\ 

10 

N 

/ 

^ 

' 

\ 

^ 

-^ 

lJ 

L— 

— ' 

- 

— 

^^1   V^ 

— ' 

W 

- 

L5 

(Dai-H 

y 

=7 

/ 

(d)a:i- 

V 

•=M 

1 

/ 

1 

1 

1 

_ 

_ 

Fig.  63. 

These  curves  are  seen  by  the  figure  to  intersect   in  four  points, 
which,  if  the  figure  is  carefully  drawn,  are  seen  to  be  about 


r  X  =  2  r  X  =  - 1.8  r  X  =  3.1  r  x  =  -  3.3 

The  results  for  B.  C,  D  are  not,  of  course,  precisely  accurate;  the 
answers  can  be  found  to  any  number  of  decimal  places,  however,  by 
drawing  the  figure  on  a  sufficiently  large  scale  on  a  large  sheet  of 
paper.     The  exact  results  to  four  places  are : 

rx=2  rx=- 1.8481  |x=.3.1313  rx=- 3.2832 

^■[^  =  3'     ^'  I  y= +3.5844'       '[^  =  -2.80.31'        '{^=-3.7794' 


278 
Ex.2. 


SIMULTANEOUS   EQUATIONS 


[Ch.  X 
(1) 


A 

^ 

^ 

r 

— 

— 

/ 

(a)/ 

\ 

1 

\ 

0 

J 

f 

V 

y 

\ 

/ 

\ 

/ 

V 

L 

U" 

1 

Fig.  (i4. 


The  figure  is  as  shown  ('Pig.  64);    (1)  is  a  circle  of  radius  5,  (2)  is 
the  cubic  curve  drawn  on  p.  189.     A  table  of  values  for  (2)  is : 


X 

0 

0 

1 
1 

2 

8 

3 

27 

4 
64 

etc. 

-3 

5 

7 

etc. 

-  1 

_  2 

etc. 

-27 

IQPi         '^/IQ 

1         ^ 

y 

Ex.  3. 


The  solutions  are  seen  to  be  abovit  (x  =  1.7,  y  = 'i.'i)  and 
(x  —  —  1.7,  y  ——  4.7).  They  can  be  found  more  accurately  f^-om  a 
larger  figure. 

{y  =  ^\  (1) 

\y  =  hx-1.  (2) 

Equation  (1)  is  the  cubic  curve  drawn  above. 

Equation  (2)  is  a  straight  line  through  {x  =  0,  y  =  -2)  and  {x  =  5, 
y  =  23). 

The  figure  is  as  shown  (Fig.  65),  with  2^  small  si)aces  as  unit  ver- 
tically, and  one  large  space  (5  snuiU  spaces)  as  unit  horizontally;  this 
is  convenient,  as  will  be  seen  by  trying  to  draw  the  figure.  The  solu- 
tions are  (approximately)  : 

r  X  =  2 1  r  X  r=  .4 1  r  X  =  -  2.4 

A:\  i;      B:\  . 

[y  =  HJ  U  =  .l 


C: 


14 


\M] 


SLML'LTANEOUS    QUADRATICS 


279 


It  is  iuterestiug  to  notice 
that  we  have  really  solved  the 
equation 

(3)  x^  -  5  X  +  2  =  0. 

For,  if  we  subtract  the  sides  of 
(•J)  from  those  of  (1)  respec- 
tively, we  get  equation  (3). 
Hence,  the  values  of  x  just  found 
are  the  solutions  of  (3).  (Com- 
pare p.  261.) 

Another  method  of  solving 
(3)  will  also  be  given.  We  note 
by  trial  that  x  =  2  is  a  solution. 
Hence,  x  —  2  is  a  factor  of  the 
left  side  (see  §  117,  p.  223,  and 
Appendix,  §  5).  Actually  divid- 
ing x^  —  5x  +  2  by  x  —  2  we  find, 
x^-5x  +  2=(x-2)(x^  +  2x-l), 

hence,  (3)  is  the  same  as 
(x-2)(x-'  +  2x~l)=0, 

whence,  either 
X-  -  2  =  0,  or  x^  +  2  X  -  1  =  0. 

That  is,  either 

X  =  2,  or  X  =  —  1  ±  ^2. 
Since      V2  =  1.4142  ..., 

the  answers  are 

X  =  2,  or  X  =  +  .4142 


:  :    :      _qr  j 

dt  t 

j:. 

±7 

__  ±: 

tl 

tt 

t 

H> 

ft 

._x 

tA- 

i^ 

'             -,1 

, 

_t : - 

n 

^  t 

-  -      J  - 

._:_  .L, - 

■4     J 

^i" 

/l/ 

-  ■    .- 

/ 

/  X 

/  / 

J  i^ 

1 

f]l^^       2-         3 

i,t                  "2                      "                           ' 

>- 

'    l 

-J- -j4--j   --T-I-- 

I                         '- 

t                              \ 

.   "^    !  ;  1 

d'        1 

JJ_SCAL2:Ul          1 

J        -.- 

•        1 

c       t 

\        i  - 

^ 

n3~ 

i    ~ 

t    . 

-6                    __... 

1 

i       i     - 

t       ' 

ltd-     J 

t 

=8 

i 

;       '^i 

n9 

t     J 

i 

-  jj )         

1             : 

1 

nil 

1 

X 

t         -    - 

/  / 

313 

ifi 

1 

~y  1 

-15   - 

3^      It 

~r 

/   ' 

- 1,5 

/  f  1  1 

_L 

Lil  1 

1 

.Y 


Fig.  65. 


orx=  -  2.4142 -, 
which  are  more  accurate  values  than  those  found  by  our  figure. 


Subtracting  these  values  in  (2),  we  find 

y=8,ory=  +  0.0710. •.,  or  .y  =  -  14.0710...; 


lience,  the  more  precise  answers  for  the  simultaneous  equations  (1) 
and  (2)  are : 


.4: 


r  X  =  2 


B: 


X  =  .4142 

//  =  .0710 


C: 


r.*' 


2.4142 


'  [i/=  -  14.07 


10 


280  SIMULTANEOUS   EQUATIONS  [Ch.  X 

EXERCISES    VI:    CHAPTER   X 

Solve  graphically : 
^^     ^.1-^4-3^  =  7,  g_     |x^  +  2/^  =  100, 


2. 


f  ah  =  21,  ^      r  m7i  =  10, 

y  +  z  =  2,  p  =  <3, 

8. 


z--y-=  16.  [2  =  ^-1-6. 

x-  +  f  =  25,  9^    ^^_jx-5  =  0. 


5. 


xy  =  12. 

2r-\-pq  =  30, 

p  =  5q\  11.    x^-x^-2  =  0. 


10.    v/^-.">?/  +  2  =  0. 


REVIEW  EXERCISES  VII :   CHAPTER  X 

Solve  the  following,  graphically  and  analytically 
2  =  A;-  +  7,  n      i  •''  =  —  5  ^'> 

Z"  " 


2. 


25fc-  =  31.  [x'-\-7y^  =  36. 

(x  +  y  +  xy=:13,  q.     f  ■^•  +  2/  +  ^'  =  4, 

,  ,     ,  ,  .0  \x^  +  4xy  +  2y^  =  17. 


3         P  +  Zl,u  =  10,  9.      ^^,^^j2. 


?»i  —  m^  =  1. 


n-  +  uv  +  v^  =  l,  ■^°-     1  ^,^2  _  ^,^,^  ^  j5^ 


»H?i  —  n"^  =  6, 


2v2_3„2==5, 


a''-l/  =  32, 


7-2  +  s^  =  45,  ^^'     I  (a  -3)^6'^  =  13. 


S        Z 


12. 


(«-5)^'+(&-3)-  =  5. 


?y-  =  5  re,  _^^      r  x-  +  ?/-  n'-  a-  +  y  =  98, 

a-^  +  7  2/^  =  36.  ■     \x--y-  +  x-y=14:. 


1:31]  REVIEW  281 

Solve  graphically: 

14      |ar'+?/  =  54,  16.    ar^ -39x- 70  =  0. 

17.    .T'*-9a^-4a;  +  12=  0. 
ar  —  y-  =  15, 

xy  =  m.  18.    s'-s2_4s  +  5  =  o. 


15. 


19.  A  24:-foot  rope  is  exactly  long  enough  to  surround  a 
right  triangle,  whose  hypotenuse  is  10  feet  long.  How  long 
are  the  other  sides  ? 

20.  The  diagonal  of  a  rectangle  whose  area  is  60  square 
inches  is  13  inches  long.  What  are  the  length  and  breadth  of 
the  rectangle  ? 

21.  I  find  that  I  can  walk  half  a  mile  farther  in  an  hour 
than  John  ;  also  it  takes  me  half  an  hour  less  to  walk  14  miles ; 
how  fast  do  we  each  walk  ? 

22.  The  radius  of  curvature  of  a  concave  mirror  is  15  cm. 
How  far  from  the  mirror  must  au  object  be  placed  in  order 
that  its  image  be  distant  20  cm.  farther  ?     (See  p.  220.) 

23.  A  loop  of  twine  30  inches  long  is  to  be  stretched  over 
four  pegs  so  as  to  form  a  rectangle  whose  area  shall  be  44 
square  inches.     What  must  be  the  sides  of  the  rectangle  ? 

24.  The  radius  of  curvature  of  a  concave  mirror  is  r.  AYhere 
must  an  object  be  placed  in  order  that  the  center  of  the  mirror 
shall  lie  halfway  between  the  object  and  image  ? 

(In  solving  graphically,  choose  any  convenient  value  for  r.) 

25.  Find  two  numbers  whose  sum  is  27,  and  whose  product 
is  126.     (See  also  p.  224.) 

26.  Find  two  numbers  whose  difference  is  19,  and  whose 
product  is  216. 

27.  In  order  that  an  object  20  cm.  farther  from  a  concave 
mirror  than  its  center  may  produce  an  image  30  cm.  from  the 
mirror,  what  radius  of  curvature  must  be  chosen,  and  where 
must  the  object  be  placed  ? 


282  SIMULTANEOUS   EQUATIONS  [Cii.  X 

28.  A  can  fold  3000  advertising  circulars  in  three  hours  less 
time  than  B.  The  two  working  together  can  fold  7500  of 
them  in  five  hours.     How  many  can  each  fold  in  one  hour? 

If  there  are  n  numbers,  the  first  of  which  is  a  and  tlie  last  /,  such 
that  the  differences  between  consecutive  pairs  are  all  equal  to  the 
same  number  d,  then  it  is  known  (see  Chapter  XIII,  p.  324,  §  157) 
that 

I  =  a  +  (»  —  l)c?, 

and  that,  if  s  represent  the  sum  of  the  n  numbers, 

s^-(a  +  l). 
2 

29.  A  row  of  numbers  is  written  down,  beginning  with  3, 
such  that  each  number  is  obtained  from  the  preceding  by  add- 
ing 4.     If  the  sum  of  the  numbers  is  55,  find  all  the  numbers. 

Here  a,  d,  and  s  are  known  ;  we  therefore  have  one  linear  and  one 
quadratic  equation  to  determine  n  and  I.  When  n  is  known,  all  the 
numbers  can  at  once  be  written  down. 

30.  The  sum  of  all  the  integers  from  100  up  to  a  certain 
integer  is  2800.     What  is  this  integer  ? 

31.  The  sum  of  all  the  odd  integers  from  1  to  a  certain  num- 
ber is  324.     What  is  the  last  odd  number  of  the  list  ? 


134]  SUMMARY  283 


SUMMARY   OF   CHAPTER   X:     SIMULTANEOUS   EQUATIONS 
INVOLVING   QUADRATICS,     pp.    253-282 

Pakt  I.     OxK  LiNE.\R  AND  Onk  QrAUKATic.     pp.  253-266. 

Figure  for  one.  Linear  ami  one  Quadratic  :  straight  line  aud  curve 
as  in  Cliapter  IX. 

Answers:  pairs  of  values  at  points  of  intersection;  approximate 
answers  from  figure;  check  on  algebraic  solution. 

§  127,  pp.  253-255. 

Definitions :  degree  found  by  adding  exponents  of  unknowns. 

§  128,  p.  255. 

Formal  Rule  for  Algebraic  Solution  :  essentially,  solve  the  linear 
equation  for  one  letter  and  substitute  this  value  in  the  quad- 
ratic.    Exercises  I  (§§  127-129).  §  129,  pp.  255-260. 

Second  Graphical  Method,  Ordinary  Quadratic :  equivalence  of  roots 

of  x^  +  px  +  (J  =  0  to  roots  of  J  ^  ~  I  ;  list  of  exam- 

[y  =  -  px  -  q\ 

pies  in  one  figure.     Exercises  II.  §  130,  pp.  261-265. 

Review  Exercises  for  Part  1,  Chapter  X  :  Exercises  III. 

pp.  265-266. 

Part  II.     Simultaneous  Quadratics,     pp.  267-282. 

Simultaneous  Quadratics  :  Figure  —  two  curves;  answers  —  pairs  of 
values  at  points  of  intersection;  algebraic  solution  illustrated 
—  methods  of  Chapter  VI.    Exercises  IV.      §  131,  pp.  267-270. 

Possibilities :  possibility  of  no  answers ;  effect  of  raising  a  curve 
vertically ;  all  cases  shown  —  no  answers  up  to  four  sets  of  an- 
swers.    Exercises  V.  §§  132-133,  pp.  270-276. 

Necessity  of  Graphical  Solution  ;  failure  of  algebraic  methods;  re- 
course to  approximate  graphical  solution.     Exercises  VI. 

§  134,  pp.  276-280. 

Review  Exercises  for  Part  II,  Chapter  X :  Exercises  VII. 

pp.  281-282. 


CHAPTER  XL     RADICALS;    FRACTIONAL  AND 
NEGATIVE    EXPONENTS 

PART    I.     OPERATIONS;   FRACTIONAL   AND 
NEGATIVE    EXPONENTS 

135.  Essential  Rules.  We  have  seen  (§  105)  that  roots 
may  be  operated  upon  in  tlie  fractional  exponent  notation 
(§  102)  whenever  the  roots  can  be  found  otherwise. 

The  student  should  read  again  and  state  the  definitions  of  §  94, 
p.  181,  and  §  97,  p.  186;  and  he  should  review  §§  98,  99,  102,  103,  104, 
105,  with  great  care.  A  few  important  statements  are  now  repeated 
in  some  cases  in  greater  detail. 

A  rational  fraction  is  the  quotient  formed  by  dividing 
one  integer  by  another. 

A  rational  number  is  an  integer  or  a  rational  fraction. 
Any  number  that  is  not  rational  is  called  irrational. 
(§  97.)  All  rational  and  irrational  numbers  are  called 
real  numbers. 

A  radical  is  any  expressed  root,  whether  rational  or 
irrational,  and  any  expression  that  contains  a  radical  is 
called  a  radical  expression.  The  index  of  the  expressed 
root  is  the  degree  of  the  radical. 

Thus,  V2  or  2^,  \/4  or  4^,  Vi'^  —  4  ae  or  Qfi  —  ^ncy  are  radicals  of 

the  second  degree,  or  quadratic  radicals ;  v5  or  5^,  vx'^  or  (x^) »,  are 
radicals  of  the  third  degree,  or  cubic  radicals;  3  —  ^2,  2v5—  1, 
X  +  vx"'^  are  radical  expressions. 

A  surd  is  a  radical  tliat  is  irrational.  A  surd  expression 
is  one  that  contains  surds ;  such  an  expression  is  said  to 
be  surd.     Any  radical  expression  is  either  surd  or  rational. 

284 


FRACTKJNAI.    AND   NEGATIVE   EXPONENTS       285 

Thus,  of  the  radicals  mentioned  above,  VS  and  "v/5  are  surds;  V4 
is  not  a  surd  ;  vV-  and  Vb^  —  'iac  may  or  may  not  be  surds  according 
to  the  values  of  tlie  letters :  if  x  =  2,  \/x-  =  y/i  (surd)  but  if  x  =  8, 
v^=v^=4  (not   surd);    in   Chapter   VIII,    §  lU,    the   radical 

expression  — — — — —  is  used  extensively;    in  some  examples 

it  is  a  surd  expression,  in  others  it  is  not.     (See  §  114,  p.  212.) 

In  this  Chapter,  we  shall  not  deal  with  even  roots  of 
negative  numbers,  which  are  often  called  imaginary  num- 
bers. (See  §  94  and  §  113.)  The  definitions  given  above 
exclude  such  roots.      (See  Appendix,  §  31.) 

We  shall  frequently  use  the  rules  of  §  104 : 

I.    (a;"')"  =  ^"""• 
II.    x"'  xa;"  =  a;"'+". 
III.    a-»x/'  =  (a--^)";  Ilia.   ^  =  ('-Y. 

These  rules  were  proved  for  simple  powers  only,  i.e. 
for  positive  integral  values  of  m  and  n.  (See,  however, 
§  105.)  Let  us  now  try  to  follow  out  the  work  already 
done  by  extending  our  notation  as  in  §  102 ;  and  in  doing 
so  let  us  observe  the  rules  1,  If,  and  III,  as  well  as  the 
rules  of  p.  35,  if  possible. 

136.  Meaning  of  Fractional  Exponents.  The  student 
already  knows  the  notation  of  §  102: 

(1)  a;^=Vrr; 

(2)  x^  =  ^,  etc. ; 

or,  in  general, 

1  _ 

(3)  x^  =  </x. 

Consider  now  (^.r)^;  if  Rule  I  is  to  hold  for  fractional 
values  of  the  exponents, 

{^xy  =  (x^y=x^''^  =  xi 


286  RADICALS  [Ch.  XI 

In  general,  by  Rule  I, 

(Vxy  =  {x'^y  =  z''''''  =x'^. 

If  Mule  I  is  to  hold  for  fractional  exponents, 

p  _ 

x'i  =  {  Vxy. 

In  this  work  we  shall  mean,  as  before,  the  positive  answer  in  case 
there  are  two,  unless  the  uegative  answer  is  esjiecially  indicated  by  the 
sign  — ,  as  in  §  94.  But  in  the  case  of  odd  roots,  where  there  is  only 
one  answer,  we  shall  mean  that  one,  whether  it  is  positive  or  negative. 
Even  roots  of  negative  numbers  are  excluded,  as  stated  in  §  135, 
throughout  this  Chapter. 

137.  Multiplication  and  Division ;  Radicals  of  same  De- 
gree. Consider  the  product  V2  •  V8 ;  this  may  be  writ- 
ten V2  .  V8  =  2^  .  3^  or,  by  Rule  III : 

V2  •  V3  =  22  .  3^'  =  (2  X  3)^  =  6^  =  V6. 
This  seems  otherwise  reasonable,  for  (  V2)2  (\/3)2=  (\/2  •  Vsy=  6, 

hence,  (V2  •  v/3)2  =  6  or  V2  •  V3  =  Ve. 

11  1  

In  general,  -y/a  Vi  =  a"  x  b"  =  {aby  =  ^ah. 

-r,;-!  ,  '\/ab  n/-  ^X  nfx 

Whence,  also,  =  Va,  or  —  =  \/-. 

Even  roots  of  negative  numbers  are  excluded  here,  as  elsewhere  in 
this  Chapter,  and  the  agreement  of  §  94  is  maintained.     See  §  136. 

Rule.  Any  root  of  a  product  of  several  factors  is  equal  to 
tlic  fvoduct  of  the  similar  roots  of  each  of  the  factors  sepa- 
rately. This  rule  enables  us  to  multiply  radicals  of  the 
same  degfree :  it  also  enables  us  to  remove  a  factor  that 
is  a  perfect  power.  Reversing  the  rule  enables  us  to 
divide  one  radical  by  another  of  the  same  degree. 

Ex.  1.    ^^3  X  a/5  =  3^  X  r»5  =  (3  .  5)^  =  15^  =  a/15. 
Ex.  1  tt.    Similarly,      ^^=a/3. 


1:30-137]  FRACTIONAL   AND   NEGATIVE   EXPONENTS    287 

Ex.  2. 

V2^V37>  =  (2  a-^)5(3  6)2  =  (2  a'  x  'A  6)'  =  (6  a'b)'  =  V6"^. 

Ex.  2  a.    Similarly,  VO  a-^6  ^  ^r^-^ 
V2a« 

CAec^- :    If  a  =  2,  ft  =  3,  V2a3  =  Vie  =  i ;   v'36  =  V9  =  3. 


V6~^  =  V6  •  8  ■  3  =  Vl44  =  12 ;  4  x  3  =  12  (correct). 

In  checking  answers  by  substituting  random  numbers  for  letters, 
if  a  little  care  is  used  in  selecting  values  of  the  letters,  the  radicals 
can  alwaj^s  be  made  exact.  Do  not  set  any  letter  equal  to  0  or  1,  for 
all  powers  of  either  of  these  numbers  are  the  same  and  errors  would 
therefore  be  concealed.  In  any  case,  such  a  check  is  not  complete ;  it 
merely  shows  that  the  woi-k  is  prnhnbly  correct. 

A'alues  of  the  letters  which  would  lead  to  even  roots  of  negative 
numbers  in  any  instance,  are  excluded,  as  stated  in  §  135. 


Ex.  3.    Vl2  =  V4  X  3  =  (4  X  3p  =  4^  x  3^  =  2  x  3^  =  2  V3. 

(See  p.  188.) 

The  factor  4,  when  removed  from  beneath  the  radical  sign,  gives  a 
factor  2  outside  it. 

Ex.  4.    Find  the  product  of  V2a^  X  V6  ab\ 

■y/2~a%  X  \/^'aU^  =  (2  a%y-  x  (6  ah'^y  =  (12  a'^b^)^- =  (4  a^V^  X  3  b)^ 

=  (4  r/4//^)i  X  (3  by  =  2  (fib  X  (3  by,  or  2  a%  VJU- 

Here  the  factors,  4,  a^,  ft^,  are  each  perfect  squares ;  they  give  the 
factors  2,  d^,  b  when  taken  outside  the  radical  sign. 

Check :     Let  a  =  2,  ft  =  3  ;  then  the  problem  and  result  become 

(2  •  8  •  3)2(6  •  2  •  9)2  =  4  .  2  .  3  V3T3,  or,  (48)^(108)^  =  24  •  3, 
or  (48  X  108)2  =  72,  or  (.5184)2  =  70  (correct). 

The  real  justification  for  the  work  done  above  consists  in  the  fact 
that  any  surd  radical  can  be  expressed  as  nearly  as  we  please  by  a 
decimal  fraction,  and  the  products  of  these  approximate  values  of  the 
radicals  follow  the  rules  of  §  104  because  they  are  exact  roots.  (See 
§  105.)  A  further  explanation  of  these  underlying  facts  is  given  in 
the  Appendix,  §  30.  Just  now  we  shall  be  satisfied  with  the  argu- 
ments given  above,  wliicli  .sliow  that  these  operations  are  I'easonable. 


288  RADICALS  [Ch.  XI 

Ex.  5.    One  reverse  of  Ex.  4  is  2  a-h  Vo  6  -=-  V2  a?h. 

The  multiplication  of  binomials  and  longer  expressions 
is  easily  done  by  the  rules  of  Chapter  IV  (see  §  98,  p.  187). 

Ex.6.   (2  +  V3)(3-2V2). 

(2  +  \/3)  (3  -  2  V2)  =  2  (8  -  2  V^)  +  V3  (3  -  2  V2) 
=  6  -  4  V2  +  3  V3  -  2  V6. 

EXERCISES    I  :    CHAPTER   XI 

In  the  following  exercises  perform  first  the  operations  indi- 
cated, if  any;  then  reduce  to  as  simple  a  form  as  possible, 
removing  all  possible  factors  outside  the  radical : 


V50.  5.    ^64  a^ft^V. 

V252.  6.    a/64  a'-'6'V. 

V300  a^b^c*.  7.    ^10.^4. 


i/ASxYz\  8.    V3x'yW4.xif 

9.    i/Wxy\?b  '  -y/B  d'yz  •  -^6  az^. 
10.    VlO  ax^i)  '  VS  a^yz  •  V6  az  . 

11.  (V3^)-^  12.  i^/soLy. 


13.    V(2;  -  x)(x  -  y)  •  V(a;  -  y){y  -  z)  •  V(?/  -  z)iz  -  x). 


14    _4^.  17  (^-.T)sy(.T-yy(y-^y 

</3a%*  '  ■V{y-z){z-x){x-y) 

V3^  "  (36  +  c)V&-c 

^5  m'n^p^  20.  (5-3V6/. 


l;!7-l;iS]    FRACTIONAL  AND  NECJATIVE  KXI'ONKNTS      :i89 

21.  (7-V5)l  25.  (x  +  y  Vz){a:  y/z  -  y). 

22.  (V7-V2)l  26.  (2  +  ^4)(^2-3). 

23.  (3  +  2  Vr))(2  -  3  V3).  27.  (6  +  V5)(l+V2). 

24.  (3-^2)^  28.  (2-V2y^  +  (3  +  V2)l 

29.  (2  +  V7)(l-3V7)(5V7-19). 

30.  (6-V3  +  V2)— (1  +  2V3-3V2)2. 

138.  Addition;  Similar  Radicals.  Radicals  can  be  added 
together  (or  subtracted)  only  when  they  involve  the  same 
root  of  the  savie  number. 

Thus,  2  VS  +  5  V;5  =  7  V;] ;  but  "2  \/3  +  ;i  Vo  must  remain  as  it 
is.  We  can  in  any  case,  of  cour.se,  find  approximately  the  value  of 
the  radicals,  and  add  these  approximate  values.  Thus,  V.3  =  1.7;)-i 
(approximately)  and  VS  =  2.236  (approximately) ;  hence,2  Va  +  3  Vo 
=  10.172  (approximately). 

Radical  monomials  whose  radical  factors  are  the  same 
roots  of  the  same  quantities  are  called  similar,  otherwise 
they  are  dissimilar. 

By  §  137  many  forms  which  appear  dissimilar  may  be 
made  similar.     They  may  then  be  added. 

Ex.  1.    Add  7  V3  +  Vl2  -  (75)1 

7V'3  +  VT2  -  (75)^  =  7  V'3  +  \/4  x  3  -  V2b  x  3 
=  7\/.3  +  2a/3  -  5 V3  =  4 Vs. 

EXERCISES   II:   CHAPTER   XI 

Simplify  the  following  expressions: 

1.  3V98-V50-f-2V8.  4.   4^54-^^250-^16. 

2.  V300-3V27  +  5VI2:      5.    (256)' +2(4)^-2(108)1 

3.  2(54)^  +  4("J16)i-(150)2.     6.    7V45-2(20)i  +  V2000. 


290  RADICALS  [Ch.  XI 


7.    Va-/A-^-V^.f-/  +  V6.  9.    ^54a.-2-V16a,-'  +  -v/'250ir^. 


8.    6V8A;-V32A;+(162)U-.    10.    V-i-'r -f  V27  2/^-(64 /z^)3. 


11.    VC-v"  -  7  a;  -f  6)  (a;  -  1)  +  V4  x  -  24. 


12.  Var'  +  3  ;c^  +  3  ic  +  1  —  Var'  +  x''^. 

13.  ■\/x^  -  6  x^  +  9  a;  -  Vip'  +  V9^. 

139.    Reduction  to  Different  Degree.     Consider   the   ex- 
ample V25.     We  have 

or,  by  Rule  I,       _  i         ■,         i         _ 

a/25  =  (52)'  =  5*  =  5^  =  V5, 

which  is  correct  if  Rule  I  is  to  hold. 

p  p-n 

In  general,  (V.c")  =  x' =  x"-"  =  "^x^", 
which  enables  us  to  reduce  radicals  to  different  degrees. 

This  rule  will  appear  more  reasonable  if  we  notice  that,  on  raising 
p  pn  I    p\''"'         n^yn 


each  side  of  a;»  =  x»"  to  the  power  qn,  we  find  \£'i]  x=^'^"'  or 
xP»  =  x*""  by  Rule  I ;  but  the  last  result  is  surely  correct.  If  we  agree 
to  take  the  positive  root  when  there  are  two  possible  values,  we  may  re- 
turn from  the  last  result  to  the  preceding  rule,  which  is  then  justified. 

By  this  new  rule  we  can  reduce  radicals  of  different 
degrees  to  radicals  of  the  same  degree. 

Ex.  1.    Multiply  V3  by  ^2. 

V3  =  3-^  =  3«,  and  V2  =  2*  =  2« ; 

hence,  V3  x  v^S  =  3^  x  2^  =  Q^^y  x  (2-'^)^  =  (27  x  4)6 

=  (108)^  or   v/i08. 


l;}8-l:39]   FRACTIONAL  AM)  NEGATIVE  EXPONENTS    291 


Ex.  2.     Multiply  V3aP  by  Vo^. 

and  V;;^^  =  („%)  2  =  («•'/,)  «  =  («963) i. 

I  lence,  \/3al)^  x  Vo^i  =  (0  r/2/>4-)  A  ^  (a«i3)  I 

=  (9  «"/>■)  «  =  (n«i6  X  9  a^i^) 

=  (a%^)^x  (9a5/y)« 

=  a/y(9  o5^)  ^  or  «6^9a56. 


Ex.  3.     Simplify 


V4«^6^-Vl2a6« 
^^'2a 


a/2  a  V2n  Vl>  a  (2 «)  ^  (2n)^ 

=  <^^'^"'^')''  -  (6  P)  ^  =  (2  flS//)  5  _  (6  6)  ^  (62)  2 

(8  a3)  ^ 
=  (2a5J*)^-6(6  6)^,  or  v^"a¥*  -  6\/66. 

The  general  rule  in  such  cases  is  to  reduce  all  radicals 
to  fractional  exponents,  and  then  operate  as  if  the  ex- 
ponents were  integers,  by  the  rules  of  §  lO-l,  which  are 
repeated  in  §  135,  p.  285. 


EXERCISES   III:    CHAPTER   XI 

Perform  the  operations  indicated;  simplify; 

1.  V5.^7. 

2.  ^O.a/I8. 


4. 


5.   VT  d'x  ■  V21  aa?. 


6.  V(:c  —  y){z  —  xf  •  V(z  —  x){x  —  yf. 


7. 


-wab-c 


X}fZ 


V54 


V3  x-ya^ 


292  RADICALS  [Cii.  XI 

11.  (^2-V3)2. 


V6ic«3- 

-^Wx'yf 

^ 

/2tx 

-\/mnt- 

-  ^mnt 

10. 


12.  (V5-V2)(V2-V5). 
^/nmt  13.  (V'2-^3)(V2  +  ^.3). 

14.  (^12-V2)(^44+2  +  ^4608).  15.  (V3-a/2)3. 


140.  Rationalization  of  Parts.  It  is  very  inconvenient 
to  have  a  radical  in  the  denominator  of  a  fraction,  since  in 
calculating  the  value  of  the  fraction  the  denominator 
is  then  a  long  decimal,  by  which  it  is  necessary  to  divide 
to  get  the  value  of  the  fraction. 

Other  parts  of  an  expression  also  occasionally  require 
simplification  for  special  purposes. 

The  process  of  rationalization,  so  called,  is  any  justifi- 
able process  that  frees  some  part  of  a  radical  expression 
of  the  radicals  contained  in  it.  It  is  understood  that  any 
operation  is '  justifiable  only  if  it  leaves  the  value  of  the 
given  expression  unaltered. 

141.  Rationalization  ;  Monomial  Denominator.  By  far 
the  most  important  is  the  rationalization  of  monomial 
denominators. 

Ex  1      1    ^   1    V3^      V3   _V3^1V.^ 

V3      V3V3      V3V3       3        3 

As  in  this  example,  any  monomial  denominator  can  be 
rationalized  by  multiplying  both  numerator  and  denomi- 
nator a  sufficient  number  of  times  by  the  radical  contained 
in  tlie  denominator,  —  a  process  which  does  not  alter  the 
value  of  the  expression. 

Occasionally  a  simpler  multiplier  may  be  selected  by  inspection, 
the  purpose  being  to  make  the  radical  in  the  denominator  exact. 
This  will  be  learned  by  practice. 


1^9-141]    FRACTIONAL  AND  NEGATIVE   EXPONENTS    293 
^,     r)     V3T     V3  6  X  ^s/5^i  __  Vlo  ah 

I_jX,  w,       ■  ^^^  -^^^^ —  * 

V  5  a      Vo  a  x  Vo  «  ^  ^ 

/-..     7     T    i.  -    ,       o     i^i        V3T     3  ,    y/lb  ab      15      3 

(heck:  Let  a  =  :j,  6  =  3  ;  then  — ^^  =  -,   and   =  —  =  -  • 

Voa      5  5  a         25      5 


Ex.3.    -l,  =  l^=L>^=^  =  (^=^,ovlV25. 
^o     5^     5^  X  5^     53        ^  ^  ^ 

^5  =  1.7+,  and</-2o  =  2.92+  ;  hence,  -^  =  0.5S+,  and  ^-^  =  0.58+. 


j,^  ^     V^  _^ 4^ ^ 4^  2t  ^  16^  •  8^ ^  (16  .  8)^ ^  (2^-2»)^ 

■    ■    V2     2^     2»'.2^  2  2  2 

_(2^_(2«2)*_2(2)^_^i         6/^ 
~    2    ~     2     ~     2     --'"^^- 

C//et•^•.•   ^  =  1^^  =  L12+,  and  ^2  =  </V^  =  </lA]+  =  L12+. 
i/2      1-41+ 

Ex  5     ^^  =  (»&)^  ^  (ab)^  ^  (ab)^  x  a^ft"  ^  (a^6'^)gx(a^6^^ 

^/    Ta        ^Toi  1,2  1,2  21  3.3 

"     a&     -      aft      -      &     '^^jVad^ 

CAeci-;  Let  a  =  i  =  2 ;   ^^  =  ?  =  1,  and  ^  ^^^=  1  •  2  =  L 
S^p     2  b  2 


In  general,  _A_  =  ^_x  C  «  ^(7  «       ^VC-- 


BC"  x  C  "        56^" 


^C 


A  radical  expression  is  said  to  be  in  its  simplest  form 
when  there  are  no  radicals  in  any  denominator  and  when 
each  radical  in  the  numerator  contains  no  factor  that 
miofht  be  taken  out. 


294  RADICALS  [Ch.  XI 

EXERCISES    IV:    CHAPTER   XI 

Rationalize  the  denominators  of  the  following;  check  the 
numerical  ones  by  calculating  the  value  to  two  or  three  deci- 
mal places  before  and  after  rationalization : 


1. 


J_.  g     V6A;z  >^     JL.  ^       Q>ah 


V2  ■    V7^.  a/3  Vl8a 


1  4     W5y'z_         f.        xyz  ^        a;--v^ 


-^2  V2^'  V^W  -VW^y) 

a^  +  2  ah  +  b-  ^^      2m'  +  ^mn  +  n'^ 

A/3(a  +  6)=^  Vm  +  n  ^2  m  +  n 


xy  —  xz 
10.     ■      •    , .  12. 


-\Jx  -yy  —  z 
13. 


Vf^- 


1 


^a^  +  a  +  l- V(r-2a  +  l 


142.  Rationalization  of  Binomial  Quadratic  Denominators. 
When  the  denominator  is  not  a  monomial  it  may  be  very 
difficult  to  rationalize  it.  We  shall  treat  only  the  case  of 
simple  binomial  denominators,  involving  square  roots. 

Ex.1.     _l_==-lAi2-V31__^2-V3^2-V3. 
2  +  V3      (2  +  V3)  (2  -  V3)       4-3 

Check:  V3  =  1.732+ ; 

hence,  =  tt^t:;::-  —  0.28-, 

2  +  V3      3.732+ 

and  2  -  \/3  =  2  -  1.732+  =  0.28-. 

These  examples  result  surprisingly ;  it  is  very  clear  that  the  riual 

result  is  much  simpler  than  the  given  form. 

The  general  principle  upon  which  sucli  examples  are 
done  is  that  (.Va  —  V6)(Va  +  Vj)  =  a  —  b. 

The  radical  binomials  Vrt  -f  Vi  and  V«  —  V7>  are  called 
conjugate  if  a  and  b  are  rational.  Their  product,  a  —  J,  is 
rational. 

If  the  denominator  is  the  difference  of  two  quantities, 
one  of  which  is  a  radical,  multiply  botli  numerator  and 


141-14l>]    fractional  AND  NEGATIVE   EXPONENTS    '2^;') 

deiiomiiiator  by  the  suin  of  the  same  two  quantities ;  and 
vice  versa.  In  general,  the  proper  multiplier  is  the  con- 
jugate of  the  given  denominator. 

Ex  2      V«+V6^(Va+V6)x(Va  +  V&)_a+&+2Va6 
Va-Vb      (Va-V&)x(Va-hV6)  a-b 


then 
and 


Check :  Let  a  =  9  and  b  —  i; 

y/a+Vb  _-^  +  2  _r 


Va-Vb      3  -  - 
a  +  b  +  2Va6  ^  9  +  4  +  2-6  ^  25  ^  ^ 
a  - b  9-4  5 


Ex  3      ^'  ~  v.'/  _  C-^'  —  vV)  X  (^'  —  V.V)  _  ar  +  y  —  2  g-  Vy  _ 

»+V2/      (x+ V?/)x(aJ— Vy)  ^■^  — y 

Check:     Put  x  =  3,  ^  =  4  ;  student  carry  out. 
Take  care  not  to  alter  the  value  of  the  given  expression, 

EXERCISES    V:    CHAPTER   XI 

Rationalize  the  denominators  of  the  following;  check  the 
first  seven  by  calculating  numerical  values  to  two  decimal 
places  before  and  after  rationalization : 


1. 


V 

^-1 
1 

4. 

3  +  2V2 

5. 

1  +  V2 
3  -  2  V2 

6. 

10 

1 

V.^•  + 

v^ 

-1 

11. 

^l- 

Vl- 

m 

■VI  + 

^l- 

in 

X- 

12. 

^x^  +  y'  + 

zl 

_J_.  7  V2  +  V3. 

V3-1  ■  2-V3 

2  +  V3  g  2+ V5 

2-V3  ■  2V5-3' 

V3  +  2V2.  g  1  +  2V5 

V3-V2  ■  1_V2' 


13. 


14. 


my/ a  +  w  V6 


1  -  Vl  -  xY 

m ^^ 

if  +  y  —  Var'  +  2/^^  3/  —  V^'  —  2  ?Hic 


15. 


296  RADICALS  [Cii.  XI 

143.    Exponent  Negative  or  Zero.     It  is   convenient   to 
use    negative    numbers    as   exponents    in   the    following 
manner. 
Since,  by  Rule  II,  .  a;"*  x  a;"  =  2;"'+", 

it  follows  that  =  x"\ 

or,  on  putting  m-\-  n  =  r^  n  =  s,      —  =  x^'^-     (See  p.  76.) 

This    rule  works  perfectly,  when  r  is   greater   than  s. 

Thus,  a^  -^a^  =  a^^^  =  2;^,  (see  p.  76)  ; 

1  1        1        1-1        1,.,. 

and  x"^  ~  x^  =  x^  ^  =  x^^  which  is  correct, 

1  1         2  s         4  1  1 

P  X^        X^  •  X^         X^  '  X^        X^         X  •  X^  1 

X^        X^  '  X^  XXX 

Even  in  the  last  example  the  new  rule  is  the  easier. 
When  we  come  to  examples  in  which  s  >  r,  we  are  led 
to  write  down  negative  exponents  thus,  by  the  rule  above : 

—  =  a;3-5_.  ^-2. 

but  ^  =  1. 

x^      x^^ 

hence,  we  say  that  x~^  means   --,  in  order  that  the    rule 
above  may  remain  true. 


In  general, 

•^      —  ^,n-2n 

)y  the  rule  above  ; 

but 

2;«_  1  . 

hence,  we  say  that  x~"  means    — . 

Any  letter  with  a  negative  exponent  is  equal  to  the  recip- 
rocal of  the  expression  formed  by  chayiging  the  si^n  of  the 
exponent. 


U;j-144]    FKACTIUXAL  AND  NEGATIVE  EXPONENTS    297 
Likewise,  if  r  =  s,  we  find 

but  -  =  1; 

hence,  we  say  that  x^  =  1. 

Any  number  with  the  exponent  0  is  equal  to  1. 

In  this  stateiueiit  0*^  must  be  excepted,  for  —  has  no  meaning. 
The  expression  0''  therefore  has  no  meaning. 

144.  With  this  understanding  we  can  work  certain 
problems  more  quickly. 

Jt  is  to  be  remembered  that  x^  =x. 

Ex.  1.    ^  (Ex.  4,  p.  293). 

V2 

•^4.  4^  2  1  2  _1  2_1  1 

^  =  ^  =  2*  X  —  =  23  X  2  2  =  2^  *  =  2*.    (See  p.  293.) 

V2     22  2^ 

Ex.2,    i^^. 

i^^!^  =  4  a%^  X  a-36-2  =  4  a2-368-2     since  c"  =  1, 

=  4o-iJ  =  i*. 
a 

Ex.  3.    ^^  (see  Ex.  5,  p.  293). 

-vab^ 

1     5 

=  ah^  X  A-i  =  — ,  or  -  v^5.  (See  p.  293.) 
5  b 

It  is  a  useful  short  rule  for  the  student  to  remember 
that  any  factor  may  be  changed  from  one  term  of  a  fraction 
to  the  other  by  reversing  the  sign  of  its  exponent. 


298  RADICALS  [Ch.  XI 

EXERCISES   VI:    CHAPTER  XI 

Express  without  negative  or  zero  exponents : 
1.    a?h-^c\    ''   ■  3.   x'r^z-i.  5.    "^~'''  ^'^^ 


,      1.3  0.^2-2  2-2a;~^?/-V* 

Transfer  6,  y,  z,  p,  and  g  from  numerator  to  denominator,  oj" 
from  denominator  to  numerator : 

9  '^m'.  12.  -^-  15.  li«^&yp:!r\ 

Perform  the  operations  indicated : 

16.  5  a;2?/-V  X  3  x-^^^-*.  21.  {x-hfz'^y. 

17.  ^  _3     ,  •  22.  (a-'?>--Cir^)"l 


18.  2  a^V^!"^  X  3  a;"  V^^!^.  23.    ^  )-s-H~^. 

19.  (30r'.v-'*z^)^(5a;-7r''3b-  24.    (.t;^-?/--'/. 

20.  7a''6"Mx5a^6V-ix2(a/>r)~i        25.    (a^  +  a~^)2. 


Change   from    the  sign  V   to    fractional    exponents ;    then 
simplify  : 

2g     Va;-y  Vz  28.   — ?ii^^^. 

27.      —                 •  29.     3-              -^         ^     • 


144] 

Simplify 


RKVIKW:    PAKT    I;    OPERATIONS 
REVIEW  EXERCISES  VII :    CHAPTER   XI 


29y 


^/l08  xY'^' 

(^/2F/)^ 
9jx/  _ 


7.  (Vr+Vs)-.     13.    V432(i/-z)^a6V. 

8.  (Vr  +  V^j^     14.    (tt  — 6Vc)(c  — rfVe). 
(x  -  y)-%y  -  z)-'   . 


9.    V3a  •  V  o  & 

10.    (V7-\/3)^. 
15  a&a;'^y   _ 
-v^9  ab'^x^y- 


11 


12. 


2-ia--&"'^ 


15. 


16. 


17. 


V(a;-2/)-''(y-z)- 

xyz  —  3  2va^.v^ 

-Vxyz^ 
■V  ran  •  -Vmhi^ 


18.  V50(a=^+3  a^fe  +3 a6-+  &=').  21.    Vi<^'  •  a/6x-^/z-  ^/3  x^y. 

19.  V4^^2  •  VlSmn^p .  V3]Z  22.    ^432  +  ^l024  -  ^U58. 

20.  ^56TF.^14A^-.  %/77^    23.    3V288-V450-V578. 


24.  Vaj^-3ar^-9a;  +  27  +  3Va;  +  3. 

25.  (^/m  +  ^  -  v'^";^)(A/m  +  a/h  +  v't^. 


26.    2 Va.-^  +  ax^ - a-x -a^- Vx^ -3a^  +  3a-l- VJaT-Ta. 
Kationalize  the  denominators  of  the  following : 


27. 


28. 


xy 


3/ 2 


33 


29. 


30. 


Va_+V6. 


31. 


32. 


3-V2 

2-^3V2_ 
3  +  2V2 


34. 


Va;  +  3  —  V;f  —  3 
nVa  —  riWb  y/x  +  S  +  Vx  —  S 

Express  in  terras  of  integral  exponents  and  roots : 


35.     il^L^,  36. 


X^qr 


r2h,o 


3  m-^oA- V2  aV^: 


P  -y 


-3\-i 


ad'^c  ^ 


PART   II.     APPLICATIONS  :    RADICAL   EQUATIONS 

145.  Radical  Equations.  Equations  involving  radical 
expressions  are  usually  to  be  solved  by  clearing  of  the 
radicals,  which  can  be  done  by  raising  both  sides  of  the 
equation  to  the  same  power.  Practice  in  arranging  the 
work  will  make  it  easier  to  see  just  how  this  should  be 
done. 

In  the  first  place,  it  is  desirable  for  the  student  to  see 
that  false  answers  may  arise  which  might  be  deceptive. 

Ex.  1.    Va;  +  2  =  0. 
Transpose  2  :  Vx  =  —  2. 

Squaring  both  sides,  we  find  that  the  only  possible  solution  is : 

X  =  4. 

However,  x  =  4  does  not  satisfy  the  original  equation ;  for  if  we  put 

4  for  X,  we  get  ,_ 

^  V4  +  2  =  0, 

2  +  2  =  0, 

which  is  incorrect.  Hence, the  given  equation  lias  no  solution.  This 
is  upon  our  previous  understanding  that  Vi  means  +  2  and  not  —  2 
(see  p.  181). 

In  order  to  make  the  truth  clear,  let  us  set 

/  =  Vi-  +  2, 

where  I  means  the  left  side  in  example  1,  and  let  us  make  a  table 
of  values  of  x  and  I. 


X 

I 

0      1 
2      3 

2 
3.4 

3 

4 
4 

5 

6 

7 

8 

9      10 

-  1 

—  a 

3.7 

4.2 

5 

imag. 

imag. 

The  figure  is  then  as  drawn  (Fig.  0(5).  It  is  clear  that  the  left  side 
(/)  is  never  zero;  in  fact,  every  value  of  /  is  greater  than  2,  for  we  add 
Vx,  which  is  positive,  to  2  to  get  /;  hence,  the  given  equation  has 
no  solution. 

300 


RADICAL    EQUATIONS 


301 


— 

71 

— 

— 

— 

— 

— 

■^' 

' 

~~ 

~" 

~ 

- 

«- 

__ 

— — 

■ — 

J 

r- 

^ 

— 

7 

o 

I- 

fV 

X 

+ 

a 











Fig.  66. 


At  the  same  time,  consider  the  new  expression 

Z'  =  -  Vx  +  2 
which  results  by  taking  the  negative  square  root;  the  similar  table  is 


X 

I' 

0 
2 

1 
1 

2 

3 

i 

.5 

" 

7 

8 

9 

-  1 

—  a 

0.6 

0.3 

0 

-0.2 

-1 

imag. 

imag. 

Figure  67  shows  the  values  of  both  /  and  l' :  it  is  clear  that  tliese 
are  the  upper  and  lower  parts,  respectively,  of  one  single  curve. 


71 

"" 

■ 

j= 

^v 

X 

-1- 

r> 

'— 

,^ 

— 

— ' 

■^ 

^ 

^ 

k 

0 

*- 

~- 

■- 



— 

— 

,^ 

I' 

=\/i 

+ 

u 

_J 

FiQ.  67. 


302  RADICALS  [Ch.  XI 

Now  V  is  actually  zero  when  x  =  4 ;  it  is  then  this  second  part  of 
the  figure  that  really  corresponds  to  the  solution  given  above. 

In  fact,  the  ojieration  of  squarinr/  both  sides,  wliich  we  performed, 
has  just  the  effect  of  bringing  in  the  dotted  portion  of  the  figure  ;  for 
when  we  squared  both  sides  we  got 

X  =  i. 

If  we  now  ti'y  tn  rjo  hack  In  the  prrceding  equation,  we  should  take 
the  square  root  of  both  sides ;  this  gives 

\/x  =  ±  2, 

where  the  sign  +  has  equal  right  with  the  sign  — .     Thus,  instead  of 

the  original  equation  y- 

Vx  =  —  2, 

we  find  also  a  new  one,  Vx  =  +  2, 

or,  Vx  -  2  =  0, 

whose  left  side  is  precisely  I'  above. 

Squaring  both  sides  of  an  equation  is  a  process  that 
cannot  always  be  reversed ;  hence,  as  above,  the  result  of 
doing  so  may  lead  to  an  incorrect  answer  because  in  con- 
nection with  the  original  equation  a  new  one  is  uncon- 
sciously introduced. 

A  sample  of  reasoning  that  is  not  reversible  is  the  following : 

"  This  object  is  a  horse, 
Hence,  this  object  is  an  animal." 

An  attempt  to  reverse  this  leads  to  the  absurdity : 

"  This  object  is  an  animal, 
Hence,  this  object  is  a  horse." 

Such  irreversible  reasoning  is  dangerous  in  solving  equations  in 
algebra.  Care  should  be  taken  to  avoid  such  a  process  or  where 
it  is  unavoidable  to  check  the  solution  with  care.  While  we  have  not 
before  used  any  irreversible  process,  we  have  guarded  against  any 
unconscious  errors  by  always  checking  the  results. 


14r)-14f>]  RADICAL    EQUATIONS  803 

146.    Methods.     The  work  to  be  done  in  solving  simple 
radical  equations  is  illustrated  below. 

Ex.1.    Vl^'-  -f  i  +  7  =  X. 
Transpose  7  :  V'2  :?;+!=  x  —  7. 

Square  both  sides:  2  .r  +  1  =  x-  —  1-t  x  +  49. 

Transpose  as  indicated:        z^  —  16  r  +  48  =  0. 
Complete  the  square  :  x-  —  16  x  +  64  =  16. 

Take  the  square  roots  :  a;  —  8  =  ±  4  ; 

iience,  the  only  possible  solutions  are        a:  =  8  ±  4  =  +  4  or  +  12. 


Check:    .r  =  +  12  gives  V2-12  +  l+7  =  12  (correct), 
hence,  a;  =  +  12  is  a  solution. 


X  —  +  i  gives  V2  -4  +  1+7  =  4  (incorrect), 
hence,  x  =  +  4  is  7iot  a  solution. 

Conclusion :  the  only  solution  of  the  given  equation  is  x  =  +  12. 

Notice  that  x  =  +  4  is  a  solution  of  the  equation 


-  \/2  X  +  1  +  7  =  X, 


which  is  formed  by  taking  the  negative  radical  —  V2  x  +  1  in  place  of 
the  positive  radical  given. 


Ex.2.    V2a;  +  5- Va;-1  =  2. 
Transpose  Vx  —  1  and  s(iuare  both  sides  : 


I  X  +  .5  =  4  +  (x  -  1)  +  4Vx  -  1. 


Transjiose  all  except  4Vx  —  1  to  one  side 


4Vx  -  1  =  X  +  2. 
Square  both  sides :  16  (x  -  1)  =  x-  +  4  x  +  4. 

Transpose  as  shown :  x-  —  12  x  =  —  20. 

Complete  the  square  :  x-  —  12  x  +  86  =  16. 

Take  the  square  roots  :  .x  —  6  =  ±  4, 

or,  X  =  6  ±  4  =  +  10  or  +  2. 

Check:  x  =  +  10  ;   V2  -10  +  5  -  VlO  -  1  =  2  (correct). 


X  =  +  2 ;   \/2  .  2  +  5  -  V2  -  1  =  2  (correct). 
Conclusion :  both  x  =  6  and  x  =  2  are  solutions. 


304  RADICALS  [Ch.  XI 

Notice  that  the  equation 


V2x+  5  +  v^^n;  =  2, 

found  by  changing  the  sign  of  one  radical,  has  no  solution  whatever, 
though  tlie  work  would  lead  to  the  same  answers  as  above  except  for 
the  check,  which  would  show  that  neither  answer  was  correct. 

The  general  directions   for   solving   radical    equations 
are : 

(1)  Place  the  most  complicated  radical  on  one  side  by 
itself;  then  square  both  sides.* 

(2)  Simplify  as  far  as  possible,  and  then  repeat  step  (1) 
as  often  as  necessary  to  clear  of  radicals. 

(3)  Solve  the  resulting  equation  if  possible,  being  care- 
ful to  check  every  answer. 

(4)  Discard  every  apparent  answer  that  does  not  sat- 
isfy the  given  equation. 

EXERCISES  VIII  :  CHAPTER  XI 


1.    V2x+ 3-3=0.  9.    2a;-l+V2a;-'-7  =  12. 


2.    V3;r -8  +  2  =  0.  10.    V37)  +  l  +  l=2). 


3.    Vx-2  =  U-x.  11.   2V2z-3=  z-A. 


4.    Vx-2  =  x-14.  12.    VA;'-3A--64-8  =  2A-. 


5.    .T  +  V3  x  -  2  =  4.  13.    V2  .t2  +  5  a;  +  2  +  2  =  2. 


6.    2y  +  V/  +  7=-2.  14.    VSy'  +  y  +  2  =  i/+2. 


7.    a  +  V2  a  —  1  =  8.  15.    Vx  +  ^x—7  =  7. 


8.    3v-V?;'  +  3'y-l  =  3.         16.    Vp  +  V2p-2  =  7. 

*  In  most  of  the  exercises  below,  the  radicals  tliat  occur  are  of  the 
sern7icl  degree.  In  exercises  in  which  radicals  of  higher  degree  occur, 
raise  both  sides  to  the  corresponding  power. 


UG-147]  RADICAL   EQUATIONS  305 


17     Vi>  +  3  +  Vi>-2=5.  19.    Vx-+a;+6  =  Var'-9. 


18.    V3  s  + 1  -  V2  s  -  1  =  1.      20.    Va;+l  +  Va;-4  =  Va;+17. 


21.    V3z-2  +  V2^n-V42  +  l  =  0. 


22.    W  +  7  =  x  +  l. 


24.    Vcc  —  3  +  V.'c  +  3  =  V2  X. 


2g_    V2a;  +  5+Va;-l^^ 


23.    Va^  +  a;-Vx'  =  0.  •    ^J^^z^^-V^T^ 

147.  Square  Root  of  Surds.  The  preceding  exercises 
were  especially  selected  so  that  the  answers  should  not  be 
surd.  Otherwise  we  should  have  to  find  the  square  root 
of  a  surd  in  checking  the  problem. 


Ex.  1.    (1)   V2  x  +  4.--Vx-l=  2.    (Compare  Ex.  2,  p.  303.) 
Solving  as  above,  the  student  will  find 

X  =  7±VM  =  7  +  4V2  or  7  -  4V2. 

Check:  Vl8  +  8V->  -  Ve  +  4V2  =  2  (for  x  =  7  +  4V2), 

Vl8  -  8V2  -  Ve  -  4 V2  =  2  (for  x  =7  -  4 V2). 


We  must  then  find,  for  example,  v  6  +  4  v'2. 
The  answer  is  2  +  V2, 
for  (2  +  V2)2  =  22  +  4  V2  +  (  V2)2  =  6  +  4  V2. 

But  such  an    answer  cannot  be   found   by  inspection.     To   find    it 
directly  suppose  

(2)  Ve  +  4  V2  =  Vr  +  V,7, 

where  r  and  s  are  two  unknown    integers  or  fractions  (i.e.  rational 
numbers,  hence  not  themselves  surd,  p.  284). 
Squaring  both  sides,  we  get 

(3)  6  +  4V2  =:r +  2  V/^  +  s. 


306  RADICALS  [Ch.  XI 

If  we  can  find  values  of  r  and  s  so  that 

(4)  r  +  s  =  6, 
and 

(5)  2  Vr^  =  4  V2, 

these  values  will  satisfy  (3),  since  (3)  holds  if  (4)  and  (5)  both  hold. 
Square  both  sides  of  (4)  and  both  sides  of  (5)  ;  then  subtract ;  we  find 

(G)  r2  -  2  rs  +  s^  =  4, 

or, 

(7)  r  -  s  =  ±  2. 

Solving  (4)  and  (7)  as  linear  simultaneous  equations,  we  find 
(r  =  4,  s  =  2)  or  (r  =  2,  s  =  4)  ;  these  satisfy  (4)  and  (5),  hence  they 
satisfy  (2),  as  will  be  found  on  trial.     Substituting  in  (2),  we  have 

V  6  +  4  V'2  =  \/i  +  \/2  =  2  +  V2,  which  is  the  value  used  above. 

148.  Equality  of  Surd  Expressions.  In  §  147  we  re- 
placed equation  (3)  by  equations  (4_)  and  (5).  Let  us 
.show  tliat  this  is  surely  correct. 

Transpose  (r  +  s)  in  equation  (3) : 

[6  -  (/•  +  .s)]  +  4  >/2  =.  2  \/rs. 
Square  both  sides : 

[6  -  (r  +  s)]'^  +  2  [6  -  (r  +  s)]  •  4  V'2  +  32  =  4  rs. 

We  can  now  solve  this  equation  for  \/2  unless  its  coefficient  [6  —  (''  +  5)] 
is  zero ;  if  we  could  solve  we  should  get 

^/o^4rs-  [6-  (r  +  s)Y-'d2 
8  [6 -(,•  +  .)] 

where  the  right  side   contains  only  rational    expressions.      This   is 
iihsunl,  for  V2  is  not  rational.     (See  p.  184.)     Tlie  only  escape  from 
I  his  absurdity  is  that  the  coefficient  [U  —  (r  +  .s)]  mentioned  above 
^JH.uld  be  zero  :         6  -  (r  +  s)  =  0,  or  r  +  s^  6, 
which  is  (4)  of  §  147. 

Subtracting  this  new  equation  from  (3),  we  get 

2v^=4V2, 
which  is  (4)  of  §  147. 

This   argument   is   another   instance   of   reduction  to  an  absurdity 

(reductio  ad  al>surdum).     See  §  88,  p.  166. 

'I'liis  juslilics  conqiletely  tlie  work  of  §  147. 


117-ltS]  RADICAI.    I<:QUATI()\S  307 

In  general,  by  a  similar  argument,  we  may  say  tliat  if 

a  +  h  Vc  =  11  +  V  Vtv 

where  a,  b,  c,  ic,  w,  w  are  rational  (^i.e.  rational  fractions  or 
integers)  and  Vc  is  irrational,  then  u  =  a  and  v  Vw  =  b  Vc\ 
i.e.  the  rational  terms  are  equal,  and  the  irrational  terms 
are  equal.  It  will  be  found  that  some  surds  cannot  have 
a  square  root  of  the  form  Vr  +  Vs  where  r  and  s  are 
rational.  For  example,  the  surd  expression  \9  +  2V2 
will  give  no  rational  values  of  r  and  ,3. 

EXERCISES   IX:  CHAPTER   XI 

Extract  the  following  square  roots ;  check  by  squaring  the 
results : 

1.  V3  +  2  V2.  8.    'S/u  +  ¥^. 

2.  A/7-4V3.  9.  V40  +  10  Vi5. 

3.  ■\/l2  +  4V8.  10.  VI8  -  2  V72. 

4.  ^l  13  -  2  VH .  11.  Vt  +  2  VlO. 

5.  ^ll9  +  6  ViO.  12.  Vl8-4v'20. 


6.  A/27-6V20.  13.  Vl  +  2  Va  -  al 

7.  V16  +  GV7.  14.   V2i?-2VF^^tt2. 

15.    \2  (m-  +  ?r  +  Vm^  +  wi^^t^  +  n*). 
Solve  the  following  equations,  and  check  each  result: 


16.  V2  a;  -  6  +  V27  -  4  a-  =  3. 

17.  V6  X  +  2  +  Vl7  —  4  a;  =  5. 

18.  \/4:X-6  +  V2 .r - 3  =  1. 


308  RADICALS  [Ch.  XI 


19.    VSx-lO- Va;-4  =  2. 


20.    V3a;-10  + Va;-4  =  2. 


21.    V'2  ar^  +  2  07  +  20  -  V2  x-  + 13  =  1. 


22.    V4  .T- +  2  +  Vl2  -  x2  =  5. 


23.   -\/U  +  4:X-x^-\--VUx-x^-21  =  5. 


24.    Va^-2a;  +  28- Vic2-6a;  +  39  =  l. 


25.    V90-18a;-5ar- V34-2a:-5a;-  =  4. 

149.    Radical  Coefficients.     Equations  may  involve  radi- 
cals in  their  coefficients  only;    in  that  case  we  may  solve     • 
directly. 

Ex.1.     (2+V3)a;-4=V3. 

Transpose  4  :  (2  +  VS)      x  =  4  +  V3. 

Divide  by  2  +  V3  :  x  =  i±2^ 

^  (4  +  V3)  X  (2  -  V3)  ^  8-  2  V.3  -  3 

(2  +  V3)  X  (2  -  V3)  4-3 

=  5  -  2  V3. 
Check:  (2+ V3)(5-2V3) -4=  V3, 

or,  (10  +  V.S  -  2  X  3)  -  4  =  V3  (correct). 

Ex.2.     ar'  +  4a;-2V3  =  0. 

Transpose  2  V3 :  x^  +  4  x  =  2  VS. 

Complete  the  square  :       x2  +  4x  +  4  =  4  +  2  VS. 


Take  square  roots :  x  +  2  =  ±V4  4-2V3. 


or,  X  =  -  2  ±  V4  +  2\/3. 

V4  +  2V3  =  Vr  +  Vs. 
Then,  r  +  .s  =  4, 

r-s  =  ±2.     (See  §  148.) 


U8-U9]  RADICAL    IXiUATrONS  309 

Whence,  r  =  3  or  1  and  s  =  1  or  3. 


•\/4  +  2  V3  =  Va  +  V'l  =  1  +  V3 ; 
hence,  x  =  -  2  ±  (1  +  V3), 

or,  X  =  —  1  +  V3  or  -  3  —  V3. 

Check  for  x  =  -  1  +  V3 :  (-  1  +  V3)'^  +  4(-  1  +  V3)  -  2  V3  =  0, 

or,  4  _  o  VS  -  4  +  4  \/3  -  2  V3=  0  (correct). 

Check  for  x  =  -  3  -  V3  :  (  -  3  -  V3)2  +  4(  -  3  -  V3)  -  2  V3  =  0, 

or,  12  +  6  V3  -  12  -4  V3  -2  V3  =  0  (correct). 

EXERCISES   X:    CHAPTER   XI 

Solve  the  following  equations  : 

1.  (3  -  V2)x- - 4  =  V2.  5.  a;  +  l  =  V2(.'c-3). 

2.  (V3  +  VL>-=V6.  6    ^  +  -^"=v3 

2  -  a;2 


3.    (5  + V3)x-  +  7=3V3. 


3-a;2 


4.  4a;  +  l  =  V3(a;+2).  "    3  +  ar'      V7 

8.  x-  +  {l~  V3)  x--  2  (1  +  V3)  =  0. 

9.  a;2_5^.+  2(V2-l)=0. 

10.  a;2-9x  +  (13-3V5)  =  0. 

11.  ar'_(5_  VS)*-- V3  =  0. 

12.  a;--3V3-d;  +  4  + V6  =  0. 

13.  ar^-5V2 -x  +  S- Vl4  =  0. 

14.  a^-3(V5- V3)a;  +  16-5Vi5  =  0. 

15.  ( V2  -  1)  x-  4-  (3  V2  -  6)x  +  (2  V2  - 1)  =  0. 

16      |V3.x  +  V2..v  =  l, 
1  V2.X-V3.?/--!. 


310  RADICALS  [Ch.  XI 

17      ^.■  +  (3-V5)2/  =  4-V5, 

2a;+(l  +  2V5)^  =  2  +  3V5. 

x^  +  f  =  2Q, 
a-^  +  X  -y  =-  .>. 
(SuGGKSTiON.     Subti'act  twice  the  second  equation  from  the  first.) 

X  +  y  +xy  =  5, 


19.      , 

X-  —  y-  +  8y  =  16. 

(Suggestion.     Solve  the  second  equation  for  x  or  y,  and  substitute 
in  the  first.) 

xy  +  7  .V  =  46, 


^°'     '  7  ar^  + 10  xy  =  299. 


REVIEW   EXERCISES    XI  :    CHAPTER   XI 

Solve  the  equations : 

1.  Va;  —  7  +  V2  a-  —  7  =  8. 

2.  V7x  +  2-V3a;-2  =  2. 

3.  V a^  -  3 a; -  1  -  Vx-"  +  3ck  +  9  =  -4. 

4.  a;  —  ■\/4  ar  —  25  a-  +  6  =  6. 

5.  Vx--  -  2  a-  -  10  +  Vaf'  -  6  a;  -  6  =  6. 

6.  Va^"''  -  14  a;  +  55  -  Va.-"  +  2  a;  - 17  =  4.     , 

7.  V2ar'-2a;  +  ll-V2ar'-4a'  +  8  =  l. 

8.  V2x'2-4a--17-V2a-2-12a;  +  15  =  2. 

9.  V4a;  +  74-a^-V24a;-66-a'2  =  2. 

10.  V^M^2  +  V4a;2+2  =  3. 

11.  V6a- -  2  V8 - 2 a;  =  V48  -  10 x. 


149J        REVIEW:    PART    II;    RADICAL    EQl'ATIONS        811 


12.  Vx+2-\/2u;-10  =  V3a;-:iO. 


13.  V22-a^-\/2af'-f  1  =  6. 

14.  lx-  +  2  =  V^(ox'-G). 

15.  ar'-3x--(4  +  V())=0. 

16.  ar'-(4\/5  +  6V2)ar  +  (l4-4\/l0)=0. 

17.  (2  -  V3)  a-'  4-  (10  -  7  VS)  x  +  (17  -  9  V3)  =  0. 

18.  f^^-^- y = ^' 

12.T-3V'7  ■?/  =  2. 


19.    ^■ 


20. 


r  ( V2  +  2  V3)  .r  +  ( V2  -  V3)  .V  =  4, 
l(2V2-6)x+(3-V2)y  =  7. 

y  -  xy  =  1, 
y'^  —  4:  xy  —  3. 


312  RADICALS 


SUMMARY  OF  CHAPTER  XI:    RADICALS;  FRACTIONAL  AND 
NEGATIVE  EXPONENTS;   RADICAL  EQUATIONS,   pp.  284-311 

Part  I.     Operations  ;  Fractional  and  Negative  Exponents. 

pp.  284-299. 
Essential  Rules:   I.  (x™)"  =  x""";    II.  a,"  x  x™  =  x'»+'' ;   III.  x""  x  y" 

=  (a;  •  ^)'» ;  III  a.  x"  -i-y»  =  (  -  )  ;  basis  for  extension;  even  roots 

of  negative  numbers  —  called  imaginary  —  excluded. 

§  135,  pp.  284-285. 
p 
Meaning  of  Fractional  Exponents :  extension  under  Rule  I ;  x'  = 

(v^)J';  take  positive  answer  if  two  exist.     §  136,  pp.  285-286. 

Multiplication  and  Division,  Radicals  of  Same  Degree :  equivalence 
of  any  root  of  a  product  to  product  of   roots ;  reverses ;  divi-      \ 
sion  ;  removal  of  factors.     Exercises  I.  §  137,  pp.  286-289. 

Addition,  Similar  Radicals:  essentially,  principle  a{b-\-c)  =  ah-\-ac. 
reduction  to  similar  radicals.    Exercises  II.    §  138,  pp.  289-290. 

Reduction  to  Different  Degree :  essentially,  fractional  exponents 
obey  laws  of  fractions.     Exercises  III.  §  139,  pp.  290-292. 

Rationalization  of  Parts:  monomial  denominator  —  multipy  by  radi- 
cal in  denominator;  quadratic  binomial  denominator  —  nmltipy 
by  conjugate.     Exercises  IV  and  V.      §§  140-142,  pp.  292-295. 

Negative  or  Zero  Exponent:  x"  =  1';  x'"  =  — ;    illustrative   prob- 
lems.    Exercises  VI.  §§  143-144,  pp.  296-298. 
Reviezv  Exercises,  Part  I,  Chapter  XI :  Exercises  VII.  p.  299. 

Part  II.     Radical  Equations.  pp.  300-311. 

Generalities  :  possibility  of  elusive  answers ;  graph  ;  reversibility 
of  steps  necessary  ;  check  necessary.  §  145,  pp.  300-302. 

Methods:  clearing  of  radicals  by  throwing  radicals  to  one  side  and 
raising  to  power.     Exercises  VIII.  §  146,  pp.  303-305. 

Square  Roots  of  Quadratic  Surds  :  examples.      §  147,  pp.  305-306. 

Equidities  of  Quadratic  Swtls  :   rational   parts  equal;    surd   parts 

equal ;  justification  of  §  147.    Exercises  TX.  §  148,  pp.  306-308. 

Radical  Coefficients :  examples.    Exercises  X.    §  149,  pp.  308-310. 

Review  Exercises,  Part  II,  Chapter  XI :  Exercises  VII.  pp.  310-311. 


CHAPTER   XII 
EQUATIONS    SOLVED    BY    SUBSTITUTION 

150.  Substitution.  In  the  preceding  Chapters  we  have 
solved  a  few  examples  (see  pp.  178,  270)  by  substituting 
new  letters  in  the  place  of  inconveniently  long  expressions. 

Thus,  in  squares  of  binomials  we  used 

(X  +  ^)2  =  x2  +  2  a;?/  +  y2 

as  a   standard  (p.    08).     Given    a   more  complicated   example,  say 
(4  m^n  —  3  nmy,  we  substituted  4  m'^n  for  x  and  —  3  jnn  for  y  : 

(4  7n^n  —  3  mn)'^  =  (4  771%) ^  +  2  (4  ?«%)(—  3  in7i)  +  (—  3  nmy 

=  16  m*n  -  24  tn^n"^  +  9  nhn'^ 

by  comparison  with  the  formula  above. 

This  process,  which  we  have  already  used  so  often,  is 
called  substitution.  Apparent  answers  are  sometimes 
found  that  are  not  really  roots  (or  solutions)  of  the  given 
equations.  The  best  possible  safeguard  is  to  check  the  final 
anstvers  by  direct  substitution  in  the  given  equations. 

151.  Equations  solved  as  Linear.  Many  equations  may 
be  solved  by  substitution  more  easily  than  by  direct 
processes. 

Ex.  1.    — ^  +  3  =  5. 
Put        s  = ;    then  4  s  +  3  =  5 ;  whence,  s  =  ^, 


r2  _  O 


1  1 


x^  -  2      2 
Clear  of  fractions :  x^  —  2  =  2. 

Transpose :  x^  =  i,  or,  x  =  ±  2. 

313 


314     EQUATIONS   SOLVED   BY   SUBSTITUTION      [Ch.  XII 

4 


Check . 


srives 


4-2 


+  3  =  5  (correct). 


In  this  example  the  advantage  of  siil)stituting  a  new  letter  is  not 
very  great ;  however,  the  principle  illustrated  by  this  simple  example 
will  be  found  useful  below. 


Ex.  2. 

Put 

then, 
whence, 


2 


•4- 


10 


x-1      2y+-S 

6  5 


=  4. 


x-1      2  2/ +  3 

1  1 

x-1      "'     2y  +  3 

2  t*  +  10  y  =  3, 
+    5  t'  =  4, 

«  =  1,    V  =  4. 


(2u 

jew 


Replacing  the  letters  u  and  v  by  their  meaning  as  above, 

1      _1  1       ^1 

x-1      2'    2//  +  3      o' 

x-l  =  2,   2y  +  3  =  5; 


or, 
hence. 


Check :    Setting 


Ex.  3.    . 


x=S,  y  =  1. 

X  —  S,  y  —  \  in  the  given  equation  gives 
2  +  J/  =  3  (correct), 
64-  I  =4  (correct). 


:0, 


i  +  1: 

X     y 

i-Uo. 

U     y 


Put 
then, 


1  1 

-  =  w,  -  =  v ; 


(  u  +  t'  =  0, 
(  M  -  r  =  0, 


whence,  it  =  0  and  r  =  0  are  the  only  possible  solutions.     But  j/  =  0 
gives  -  -  0,  which   gives   no  value   for  x.     Likewise   v  =  0  gives  no 

X 

value  for  y.     Hence, the  given  equations  have  no  solutions. 


151]  EQUATIONS   SOLVED    BY   SUBSTITUTION  315 


EXERCISES   I:     CHAPTER   XII 


a;-  -  21 


+  5  =  7, 


3. 


6-2a?  l^       5-2 


x+y     x-y 

10 3 

Vx  +  y     x-y 


ar* 


=  -2. 


2  «  -(-  3  V '^ =  5, 


ou  —  zv 


0  M  +  9  y;  + 


bu-2v 


=  2G. 


10.    { 


X-  -  y^ 

3ar  +  3/ ^^=11. 

x-  -  y- 


4.    < 


^  1^     I        -^       -3 

r—s     2  r  —  5s 

.  r  —  6"      2  r  —  5  .s 


11.   -^ 


2w2-3«2  +  -— ^^ =9, 

i  r  —  3  n- 


Gn-  —  9t'  — 


13 


7«--3w2 


+  -^  =  1, 


2p  +  3fj     i^-'U 


6. 


.2/)  +  3  q     p  —  Sq 

15  7      ^8 

7-  +  s-      ?•-  —  S''      5 
50  21     ^     -^ 


.  r^  +  s^     r^  —  s' 


7.    < 


^     +.    1 


__3 

2m''-n-  '  m^-2n-     2' 

=  4. 


14       ^        3 


2m^  — 94^      m^—2)r 


12. 


r  1 


+  x-  +  /  =  62, 


50 


2x2  +  2?/2 !;:^  =  72. 

a;  +  y 


13.   < 


15  10 


45 


11. 


la^+y,2— 5     a;-Hy-5 
1 


=  1. 


14.    { 


y-4x 

25 

l?/-4a; 


-a-^  +  2/2  =  25, 
-2ar'  +  2/=73. 


x  +  y  +  — ^  =  10, 


3  a;  +  3  ^ 


10 
x  —  y 


=  17. 


15.    < 


3.r2+3a-^+3/-il^=37, 
.r  +  ^y 

X-  +  x^  +  /  H =  3( . 


316     EQUATIONS   SOLVED   BY   SUBSTITUTION      [Ch.  XII 

152.  Equations  solved  as  Quadratics.  An  equation  may 
often  be  turned  into  a  quadratic  equation  by  substitution. 

Ex.  1.  a-^-3ar^  +  2  =  0. 
Let  x2  =  s ;  then  .92  -  3  s  +  2  =  0, 

whence,  s  =:  +  1  or  s  =  +  2 

or,  x^  _  ^  ]^  or  x^  =  4-  2 ; 

whence,  x  =  ±lorx  =  ±  V2. 

Check :     For  x  =  ±  1  gives  x^  =  1,  x*  =  1 ; 

hence,  x« -3  x^ +  2  =  1-3  +  2  =  0  (correct). 
For  X  =  ±  v'2  gives  x^  =  2,  x*  =  4 ; 

hence,  x*- 3  x2 +  2  =  4-3-2  +  2  =  0  (correct). 

This  equation  therefore  has  four  roots  :  ±  1  and  ±  V2.  Draw  the 
figure  for  /  =  X*  —  3  x2  +  2  as  on  p.  204,  note  that  the  curve  crosses 
the  main  horizontal  line  at  points  where 

X  =  ±  1  and  X  =  ±  V2=  ±  1.4  (about). 

Ex.  2.  x*-x^-2  =  0. 

Let  x2  =  s;  then  52  -  s  -  2  =  0, 

whence,  s  =  2ors  =  —  1, 

or,  x^  =  2  or  X-  =  —  1 , 

whence,  x  =  ±  v'2orx  =  ±  V-l. 

Since  the  result  V—  1  is  meaningless  (imaginary,  see  p.  211)  for  the 
present,  we  say  that  there  are  only  two  answers,  x  =  ±  V2. 

Check  :  x  =  ±  V2  gives  x'^  =  2,  x*  =  4 ; 

hence,  x*  —  x^- 2  =  4  —  2  —  2  =  0  (correct). 

The  student  may  draw  a  figure  as  in  example  1. 

Ex.  3.  x^-5x'  +  i  =  0. 
Let  x8  =  s :  s^  -  .5  s  +  4  =  0, 

whence,  s  =  1  or  ,s  =  4, 

or,  x^  =  1  or  x^  =  4, 

hence,  x  =  1  or  x  =  v4. 


i 


lo2]  EQIATIONS   SOLVED   BY   SUBSTITUTION         317 

Check :     if  x  =  1,  a:«  -  5  z''  +  4  =  1  -  5  •  1  +  4  =  0  (correct). 

If  x^  =  4,  x6  =  16  ; 

hence,     x  =  \/i,  x^-5x^  +  'i-l^-5-4:-\-i  =  0  (correct). 

The  roots  1  and  Vl  found  above  are  the  only  roots  (except  imagi- 
nary roots,  wliicli  would  be  meaningless  at  present);  this  is  clearly 
seen  by  drawing  the  figure,  as  in  example  1. 

EXERCISES   II:     CHAPTER   XII 

1.  o;*  -  7  .x-2  +  10  =  0.  8.  2^-34  22  +  1  =  0. 

2.  t*-3f-10  =  0.  9.  s^_42s2  +  9  =  0. 

3.  p''_5p-  +  4  =  0.  10.  x^-10.ir  +  l=0. 

4.  4A;<-17^-  +  4  =  0.  11.  x''-2x'-\-l  =0. 

5.  9 a;'' -10a;' +  1=0.  12.  a«-9a^  +  8  =  0. 

6.  3y*-7r/  +  '2  =  0.  13.  a«  +  9a'^  +  8  =  0. 

7.  ^•<-8^•2  +  4  =  0.  14.  27c«  +  28c''+l  =  0. 

15.  82"- 192^-27  =  0. 

16.  (a;2+ 2)2 -4(^2 +  2) -12  =  0. 

18.  (ar  +  2x-  -  2)2  -  3(a;2 _^  2  .1; -  2)- 18  =  0. 

19.  2x-^-ar''-6a.-2-a;  +  2  =  0. 

Suggestion.     Since  x  =  0  is  not  a  solution,  we  may  divide  by  x^, 

with  the  result :  2x2  -  a;  -  6  -  -  +  -  =  0.    -^ow x^+ —  =  ( x +  -Y -  2. 
■,  X      x^  x^      \         xj 

The  substitution  x  +  -=  u  will  be  found  useful  in  this  case,  and  in 
x 

all  cases  where  the  coefficients  of  terms  equidistant  from  the  two  ends 

are  the  same. 

20.  3x*  +  4ar''-14ar^  +  4a;  +  3  =  0. 


318      EQUATIONS   SOLVED   BY   SUBSTITUTION      [Ch.  XII 

153.    Radical  Equations.     Equations  containing  radicals 
may  often  be  solved  easily  by  substitution. 


Ex.  1.   a;  +  3  +  ViC  +  3  =  12. 


Let  Vx  +  o  =  .s ;  then  52+5  =  12, 

whence,  s  =  +  3  or  s  =  —  4, 

or,  Vx  +  3  =  +  3  or  Vx  +  3  =  -  4,* 

whence,  on  squaring,  x  +  3  =  9  or  x  +  3  =  16, 
or,  x  =  6  or  x  =  13. 


Check:  x  =  6  gives  6  +  3  +  Ve  +  3  =  12  (correct). 

X  =  13  gives  13  +  3  +  ^13  +  3  =  20  (incorrect). 

It  follows  that  X  =  6  is  a  root,  while  x  =  13  is  not  a  root.  (See  §§  94 
and  145,  pp.  181  and  300.) 

h  is  absolutely  necessary  to  check  each  answer  in  suck  examples. 
(Compare  p.  302.) 

Ex.  2.    x^-3x^-\-2  =  0. 

Let  z^  =  s;  then  s2  _  3  s  +  2  =  0, 

whence,  s  =  +  l  or  s  =  +  2, 

1  1 

or,  x^  =  1  or  x^  =  2; 

hence, on  cubing,  x=  P  =  1  or  x  =  2^  =  8. 

Check :  x  =  l  gives  1^-3- 1^  +  2  =  1-3  +  2  =  0  (correct). 

X  =  8  gives  8^ -3-8^ +  2  =  4-3-2  +  2  =  0 (correct). 
Hence,  x  =  1  and  x  =  8  are  both  roots  of  the  given  equation. 

EXERCISES    III  :    CHAPTER   XII 


1.  a;+-V^-6  =  0.  4.    2A;  +  7-20V2fc+7+64=0. 

2.  3t  +  Vt-U  =  0.  5.    10VtH^-A/^M^-24  =  0. 

3.  z-2+7Vz^^  =  30.         6.    5Vn'  +  3  7i  +  n-  +  3H  =  14. 

*  We  might  reject  this  possibility  at  once,  since  Vx  +  3  =  —  4  contradicts 
our  agreement  that  the  sign  ^/  denotes  a  positive  answer.  The  possible 
solution  X  =  13,  which  is  derived  below  from  this,  is  seen  to  be  incorrect. 


153-loi]     EQUATIONS    SOLVED   BY   SUBSTITUTION       319 


7.  8a;t  +  19x3-27  =  0. 

8.  9h^-40«^  +  16  =  0. 

9.  9  a;' -22  x' +8  =  0. 

10.  8(/-f  i)  =  (')r)/. 


11.  4cc5_5a;o-|-l=0. 

12.  r-8V2r  +  7  +  ll  =  0. 

[Suggestion.     We  may  write 
2^4-  o-J  -  16  \/2  r  +  7  =  0,  or 


(2  r  +  7)  -  16  V2r  +  7  +  15  =  0.] 


13.  18m  +  33-llV3m  +  5  =  0. 

14.  2)^  +  1 -llVr^"^^  =  0. 


15.    X-  +  7  .r  + 1  -  4  Va^  +  7  a;  + 1  +  3  =  0. 


16.   2a-  +  3a  +  42-7V4a2  +  6a+36  =  0. 


17.  ir  +  Sj) - Ifi  +  2 Vj/  +  3p -  1  =  0. 


18.   9  z^  +  4  2  + 10  -  5  V9  z-  +  4  z  +  4  =  0. 

154.    Simultaneous   Equations.      We  can   always  solve 

simultaneous  e(iuations  of  wliicli  one  is  linear  by  the 
methods  of  §§  8<i,  129.  Simultaneous  equations  that  may 
be  solved  as  a  pair  of  simultaneous  linear  equations  have 
been  mentioned  in  §  151.  We  note  now  that  many  forms 
may  be  reduced  by  substitution  to  .one  linear  and  one 
quadratic. 


Ex.1. 


1  _1 

x"     y 
1_1 
X     y 


^  =  16, 
=  2. 


(1) 
(2) 


Let  -  =  J/,   -  =  v  ;  the  equations  become 
X  y 

w2  -  17-2  =  16, 
u  -  r  =  2, 

i     which  may  be  solved  by  the  methods  of  §  120.     We  find  m  =  5,  y  =  3; 
1     hence,  a:  =  i,  ^  =  j-     These  answers  are  checked  as  follows  : 


320      EQUATIONS   SOLVED  BY    SUBSTITUTION      [Cii.  XII 


1 


1 


Check . 


(ir  i^r' 


25-9  =  16  (correct), 


--  =  5  —  3  =  2  (correct) . 

.S  3 


It  is  obvious,  likewise,  tliat  examples  may  be  given 
that  can  be  written  in  the  form  of  a  pair  of  quadratic 
equations.  These,  however,  lead  to  rather  complicated 
work,  unless  very  special  examples  are  chosen. 


Ex.2. 


(x-2y  +  (2y-3y  =  9, 
{2y-3y  =  4.(x-2)-5. 

Write  x-2  =  u,  2y-S  =  v;  then 

W2  +  1,2  =  16, 

t;2  =  4  M  -  5. 


Subtracting  gives 
whence. 


If 


u  =  -7, 
r2  =  4  u  -  5  =  -  .33, 


and  V  =  V—  33   (imaginary) . 

Hence,    this   leads   to    no   real 
solution  ; 

(No  solutions.) 


Check : 


f  w  =  +  3, 

V  =  +  \/7 


m2  +  4  m  =  21, 
m2  +  4  u  +  4  =  25, 

u  = -2  ±5  =-7  or  +3. 
If        w  =  +  3, 

v  =  ±  y/7. 
This  gives  two  pairs  of  solutions : 

u  =  +  3, 

V  =  +  y/1. 
u2  +  f;2(=9  +  7)=  16, 
i,2(=7)=4h-5(=12-5). 


and 


u  =  +3, 

i;  =  -  \/7 . 


give 


And 


r «  =  +  3, 

|.  =  -V7     ^"^ 


Corresponding  to  these,  we  find 
x  =  u4-2  =  5,  y  ■■ 


w2+  y2(-=  9  +  7)=  16, 
r2(=7)  =  4M-5  =  (12-5). 

„  +  3  _  ±  V7  +  3 


which  the  student  should  check  by  substituting  these  values  in  the 
given  equations. 


154]         EQUATIONS   SOLVED   BY   SUBSTITUTION  821 


1.    < 


p-      q 
1_1 
P      9 


EXERCISES    IV:    CHAPTER   XII 

=  3. 


8.     i 


y      X 


-1+1  =  10. 

x^     xy 


5. 


(x-5y  +  (2y-3y  =  13, 
{x-5)-C2y-3)=:l. 

y^  +  2«  =  G5, 
y  +  ^3  =  9. 

(,.  +  s)(3r-2s)=28, 
(3  r  -  2s)- -(r  +  s)-  =  33. 


l  +  i  =  io. 


-  =  80, 


9. 


10. 


u'  +"^  =  21, 

V 

l+-  =  28. 


9     i^^      2' 
f    1     +^  =  10, 


11.  <^ 


x-y     x+y 
3  11 


12.     ^ 


X  +  ~  =  24. 

2        ^7J+3 
(m-3f  ~'/«-3' 


(x-yf     {x  +  yf 
^  —  3      a;  —  4 


=  4. 


13.     < 


{m-sy 


■  2(w  +  3)-'=-28. 


_^L ^^  =  63. 

(^-3)^      (x-Ay 

2t±t  +  x-y  =  l, 
xy 

4^±l'  +  (x--2/)^=14. 
xy 


14. 


4  ?'  —  s     V  —  1 


l(4r-s)-      (r-l)2 


322  EQUATIONS   SOLVED   BY    SLBSTITLTION 


SUMMARY  OF  CHAPTER  XII:    EQUATIONS  SOLVED  BY  SUB- 
STITUTION, pp.  313-322 

Substitution  :  new  letters  in  place  of  longer  combinations. 

§  150,  p.  313. 

Equations  solved  as  Linear:  illustrations.     Exercises  I. 

§  151,  pp.  313-315. 

Equations  solved  as  Quadratic:  illustrations.     Exercises  II. 

§  152,  pp.  316-317. 

Radical  Equations  :  illustrations.     Exercises  III. 

§  153,  pp.  318-319. 

Simultaneous  Equations  :  illustrations.     Exercises  IV. 

§  154,  pp.  319-322. 


CHAPTER   XIII.      PROGRESSIONS   OR 
SEQUENCES 

PART  I.    ARITHMETIC   SEQUENCES 

155.  Sequences  or  Progressions.  A  set  of  numbers 
tliat  follow  one  anotlier  in  a  definite  order  is  called  a 
sequence  or  a  progression.  Any  one  of  the  numbers  is 
called  a  term  or  an  element  of  the  sequence. 

In  this  Chapter  only  those  sequences  are  treated  in 
which  there  are  a  definite  number  of  terms,  so  that  the 
sequence  finally  stops.  Such  a  sequence  is  called  finite. 
Other  sequences  exist ;  for  example,  the  integers  1,  2,  o, 
4,  etc.,  •••  form  au  unending  or  infinite  sequence.  (See 
Appendix,  §§  26-28.) 

156.  Arithmetic  Progressions  or  Sequences.  A  sequence 
is  called  an  arithmetic  progression  (or  arithmetic  sequence) 
if  the  difference  between  any  term  and  the  term  which 
precedes  it  is  always  constant.  This  difference  is  called 
the  common  difference.  We  shall  use  the  word  sequence 
by  preference  from  now  on,  instead  of  progression. 

j        For  example,  the  sequence  of  integers 

1,   2,   3,   4,   5,   6,   7,   8,   9,    10 

is  an  arithmetic  sequence,  for  the  difference  between  consecutive  terms 
\  is  one  in  all  cases.     Similarly,  the  sequences 

'       (rj)       3,  5,  7,  9,      11,      13  (common  difference       2) 

I       {h)       o,  10,  15,  20  (common  difference      5) 

(c)    —4,  —1,  +2,  +.5,  +8  (common  difference       3) 

:       (d)       7,  3,  —  1,  —  5  (common  difference  —  4) 

1       (^)    ~  h  +1'  +2'  +4,  +  V  (common  difference       ?,) 
\                                                            323 


324  PROGRESSIONS  OR  SEQUENCES         [Ch.  XHI 

and,  in  general, 

(/)  a,  a  +  ft,  a  +  2  d,  etc.,  •••,  a  +  nd 

are  all  arithmetic  sequences. 

The  first  term  will  usually  be  denoted  by  a,  the  com- 
mon difference  by  c?,  as  in  example  (/). 

If  d  is  positive,  the  terms  increase ;  in  that  case  the  se- 
quence is  called  increasing.  If  c?  is  negative,  the  terms 
decrease  ;  the  sequence  is  then  called  decreasing. 

157.  Any  Term.  Any  term  may  be  found  by  adding 
to  the  first  term  the  common  difference  as  many  times  as 
there  are  intervening  steps  ;  thus,  the  second  term  is  a+c?, 
the  third  is  a  -f  2  (^,  etc. ;  in  general,  the  number  of  steps 
is  one  less  than  the  number  of  the  term :  lience  the  nth 

term  is 

a  +  (71—  l)c?, 
or,  tn=-a+  (n  —  V)d 

where  f„  means  the  nth.  term.     If  there  are  only  n  terms, 
the  nih.  is  the  last  one,  and  we  write 

(I)  l  =  a+{n-\)d, 

where  I  now  denotes  the  last  term. 

Similarly, we  may  find  the  first  term  if  we  know  the 
fourth  (say)  and  the  common  difference :  in  this  case, 
a  =  t^  —  Zd\  or,  in  general : 

a  =  tj^—  (n  —  V)d 
where  ^„  means  the  nih  term. 

Ex.  1.  If  an  arithmetic  sequence  starts  with  the  number  4 
and  has  a  common  difference  5,  find  the  10th  term. 

Here    a  =  4,  rf  =  5,  n  =  10 ;  hence,  /j^  =  4  +  9  •  5  =  49. 
Ex.  2.    a  =  —  7i,  d  =  2,  to  find  t-^^- 
Here  n  ^  12 ;  hence,  <,2  =  -  7|  +  12  •  2  =  VJ\. 

Ex.  3.   a  =  +  6,  d  =  —  4,  to  find  t^. 
/3  =  6  +  3(-4)  =  -6. 


156-158]  ARITHMETIC   SEQUENCES  325 

158.  The  Sum.  To  find  the  sum  of  all  the  terras  of  an 
arithmetic  sequence 

s=a+  (a  +  d)  +  (a-{-2d)  -\ \-  (l -  d)  +  I, 

where  I  denotes  the  last  term,  write  this  in  the  form 
sz=l-^  (l-d)  +  {l-2d)-\ \-  (a  +  d)  +a 

and  add  these  two  equations;  we  get 

2  s  =  (a  +  0  +  («  +  0  +  («  +  0  +  •••  +  («  +  0  +  («  +  0. 

the  term  (a  +  ^)  occurring  w  times  if  there  are  n  terms; 

hence,  „  ^        ,x 

2  s  =  7i(a  +  0, 

or, 

(II)  s  =  "^(a+l). 

From  I  and  II,  if  we  know  the  values  of  any  three  of 
the  quantities,  rt,  w,  d,  I,  s,  the  other  two  may  be  found. 
Thus,  given  «,  s,  d,  to  find  n  and  Z,  we  substitute  the 
value  of  I  from  I   into  II : 

s  —  -(a  +  a  +  (n  -  1)  d), 

or, 

(III)  s  =  ^{2a  +  {n-l)d), 
and  so  on. 

1      When   necessary,  to  avoid  ambiguity,   we   denote  the 
y  sum  of  71  terms  by  .s„  instead  of  s  and  write  s„  =  -  («  +  Z),  etc. 

Ex.  1.    Given  a  =  2,   I  =  14,  n  =  G,  to  find  s  and  d : 

I  becomes  14  =  2  +  (6  —  \)d,  whence  d  =  y. 

II  becomes   s  =  |(2  -|-  14),  whence  s  =  48. 

Check :  The  sequence  is  2,  4f,  G|,  9|,  11|,  14  ;  its  sum  is 
2  +  4f  +  61  +  9^+  11^  +  14  =  48  (correct). 


326  PROGRESSIONS   OR   SEQUENCES  [Ch.  XIU 

Ex.2.    Given         cZ  =  3,  n  =  4,  s  =  -l. 

I  becomes  I  =a  +  (4  —  1)3  or  /  =  r?  +  9. 

II  becomes  —!  =  *(«  +  /)  or  2  (a  +  /)  =  -  1 ; 

whence,  solving  these  simultaneous  linear  equations  for  a  and  /,  we  find 

rt  —  _19       /—If 

a  —  — 5  ,  I  —  -f'-. 
Check:  The  sequence  is 

-19-7+5+17. 
4  4  4  4 

the„.=    {^"U{-^}-.{i}  +  {!^}. -*  =  -!.  a.  give,,. 

The  general  method  is  illustrated  by  the  examples  : 
first  insert  the  given  values  in  formulae  I  and  II  ;  then 
solve  the  two  resulting  equations  for  the  two  unknown 
letters. 

Since  n  must  be  an  integer,  only  integral  values  of  n  are  given  in 
problems.  If,  in  a  problem  in  which  n  is  to  be  found,  the  value  is  not 
an  integer,  the  problem  is  impossible,  i.e.  no  arithmetic  sequence  exists 
for  that  problem. 

EXERCISES  I  :    CHAPTER  XIII 

(The  first  examples  should  be  checked  as  above  by  actually 
writing  down  the  sequence.) 
Find  the  quantities  not  given  : 


a 

d 

n 

I 

1. 

5 

3 

8 

2. 

5 

-3 

8 

3. 

i 

17 

10 

4. 

16 

-3 

-2 

5. 

-3 

1 

6. 

0 

* 

5 

7. 

-1 

7 

-4 

8. 

2 

i 

3 

-9 


158]  AKITIIMKTIC    SEQUKNCES  327 


a 

d 

9. 

15 

10. 

7 

11 

12. 

-24 

13. 

5 

14. 

6 

-1 

15. 

-1 

16. 

^ 

17. 

100 

18. 

19. 

-9 

20. 

2 

71 

19 

I 
69 

s 

5 

65 

5 

1 

-5 

3 

16 

13 

45 

-10 

7 
16 

56 
60 

300 

40,200 

6 

1 

-24 

7 

-3 

7 

21 

21.  Show  that  t,„  —  <„  =  (m  —  n)d. 

22.  Show  that  the  sum  of  all  the  terms  of  the  sequence  from 
the  ?nth  to  the  ^^th  inclusive  is  (w  —  m  + 1)^^ — ^^A — _ z_  . 

23.  Given  a  =  —  12,  d  =2,  .s  =  -  40,  find  ».  Write  out  the 
sequence,  and  indicate  why  there  are  two  vahies  of  n.  Check 
the  result  of  example  22,  by  finding  the  sum  of  all  terms  of  this 
sequence  from  the  sixth  to  the  eighth  inclusive. 

24.  Given  1  =  11,  d  =  ^,  s  =  IIG;  find  n  and  a. 

25.  Given  a  =  2,  rf  =  3,  .s  =  77;  find  n  and  I.  How  many 
true  solutions  of  the  problem  are  there  in  this  case  ? 


PART   II.     GEOMETRIC    SEQUENCES 

159.  Geometric  Sequences.  A  sequence  is  a  geometric 
sequence  if  the  ratio  of  any  term  to  the  preceding  one  is  al- 
ways the  same  ;  tliis  common  ratio  is  an  important  number. 

Thus,  1,  2,  4,  8,  16,  32,  64 

is  a  geometric  sequence,  the  common  ratio  being  2.     Also, 

3,  12,  48,  192  (common  ratio,      4), 

5,    -10,   +20,   -40,   +80  (common  ratio,  -2), 

2,  I,  |>  iV)  (common  ratio,       |), 

and,  in  general,    a,  ar,  ar%  ar^,  •••,  ar"    (common  ratio,       r). 

160.  Formulae.  As  in  the  case  of  arithmetic  sequences, 
we  may  find  two  formula? :  one  for  the  wth  term,  the  other 
for  the  sum  of  n  terms. 

If  the  first  term  is  called  a  and  the  common  ratio  r,  the 
second  term  is  ar,  the  third  is  ar^  etc.;  since  we  multiply 
by  r  at  each  step,  it  is  clear  that  the  nth  term  is 

t„  =  ar''-\ 

where  ^„  means  the  nth  term.     If  the  last  term  is  called  I, 
we  may  write 

(I)  l  =  ar"-^ 

if  there  are  just  n  terms. 
Again,  the  sum  is 

(1)  s=a  +  ar  -\-  ar"^  +  ar^  +  •  •  •  +  ar"~^  +  ar"~' . 


Multiplying  by  —  r,  we  find 

(2)     —  rs  =  —  ai —  ar"^  —  ar^  —  ...  —  ar"  ~^  —  ar^^~^  —  «r" 
whence,  adding  (1)  and  (2), 

s  —  rs  =  a  —  ar"  =  a(l  —  r"), 
328 


GEOMETRIC   SEQUENCES  329 


(II)  ^  =  ^1_. 

After  this  formula  has  been  obtained,  it  is  easy  to  see  that  it  is 
coriect,  for,  by  actual  division, 

(t^)  =  '  +  '-+^'-^  ■••  +  '■""'• 

whence,  on  multiplying  by  o, 

a  (     ~      j  =  ft  +  rtr  +  ar^  +  •••  +  a»-"~^  =  s. 

These  formula  I  and  II  may  be  used  as  were  the 
formulse  for  arithmetic  sequences. 

"Ex.  1.   Given  a  =  3,  r  =  2, 7i  =  7,  we  have 

^  =  3  .  26  =  192,       s  =  '4^—^)  =  3  (^T?)  =  3  •  127  =  381. 

Check:  The  sequence  is  3,  6,  12,  24,  48,  96,  192;  the  last  term 
is  192  (as  found)  ;  the  sum  of  the  terms  is 

3  +  6  +  12+ •••  + 192  =  381. 
Ex.  2.    Given  r  =  3,  ?i  =  4,  s  =  120. 
II  becomes   120=  "(t^?)  =  "(^-rl"^")  =  ^^^'  whence,  a  =  W  =  3- 
I  becomes  Z  =  3  •  3^  =  81. 

Check :  The  sequence  is  3,  9.  27,  81 ;  it  is  easy  to  see  that  all  the 
conditions  of  the  jiroblem,  and  the  answers,  hold  good. 

The  same  process  is  used  in  all  cases:  the  known  num- 
bers are  substituted  for  the  corresponding  letters  in  I  and 
II;  the  resulting  equations  are  then  solved  for  the  two 
unknown  quantities. 

Problems  in  which  ??  is  not  given,  but  is  to  be  found,  are  not  given 
because  the  solution  requires  logarithms  (see  Chapter  XIV) ;  these 
problems  are  also  not  particularly  valuable  since  n  is  by  its  nature  an 
wteqer;  finally,  it  is  always  easy  to  solve  such  a  problem  by  trial, 
since  only  integral  values  of  n  need  be  tried. 


330 


PROGRESSIONS   OR   SE(^UENCES        [Cii.  XI II 


EXERCISES 

II: 

CHAPTER 

XIII 

Find  the 

quantities  not  given 

a 

r 

n 

I 

s 

1. 

2 

3 

5 

2. 

16 

_  r 

2- 

6 

3. 

2 

5 

31 

4. 

4 

3 

19 

5. 

i 

5 

1 

6. 

3 

i 

4 

7. 

2 

3 

98 

8. 

3 

6 

.00003 

9. 

V3 

5 

1 

10. 

1^(1 

5 

111.11 

11. 

2 

4 

16 

12. 

V 

_2 

4 

16 

13. 

16 

5 

81 

14. 

-1 

13 

-1 

15. 

3 

2 
3 

3 

16. 

-i 

5 

H 

17. 

1 

4 

27 

18. 

3 

80 

105 

SUGGESTIO 

N.     Here 

.«  _a(i-'' 

'+r 

+  I)_r2  4- 

r+1 

19.  Show  that  t^  —  t„  =  uV-^r""-"  —  1) . 

20.  Show  that  the  sum  of  all  the  terms  of  the  sequence 
from  the  «th  to  the  mth  inclusive  is 


r-1 


1G(»-I(;i]  (iEO.METlUC   SEQUENCES  331 

161.  Theorems  in  Factoring.  The  formula  II  is  also 
useful  iu  fai;toiiiiL,s   dividing-  Ijotli  sides  of  it  by  a  : 

(1)  ^^^  =  1 +/•  +  /•-  + •••  +  /^^ 
1  —  r 

which  may  be  used  as  a  type  form  for  factoriug  expressions 

of  the  form  1  — /^,  as  in  Chapter  IV,  pp.  91-10:!. 

Again,  setting  7-  =  ^  and  clearing  of  fractions,  we  get 

X 

(2)  ^![!zJ^  =  jf/i-i  j^  x"-'^y  +x"-^y-^ \-y"-^, 

x-y 

which  is  a  type  form  for  factoring  expressions  of  the  form 
a;"  -  y". 

If  in  (1)  we  put  r=  —  .9,  we  get 

\  §n 

(8) =  1  —  s  4-  s^—  S'^+  •  •  •  —  s"~i,  if  n  is  an  even  mteger, 

1+s 

or, 

1  4-  s" 

(4)  — -  -  =  1  —  s  4-  s^  —  s"^  4-  •  •  •  4-  .s"~',  if  7i  is  an  odd  integer. 

Setting  s=-  in  these,  we  find  the  following  type  forms: 

(5)   ^=jr"-i— jr"  '^ 4- Jr"  y- —•••—/"" S  if  w  is  even, 

X  -i-y 

or, 

(6)  ^I'^Jl ^x"   '^^-x"   -y  +  x"-^y^ \-y"   i,  if  w  is  odd. 

X  -\-y 

In  English, 

from  (2)  X  —y  is  a  factor  of  x"  —y"-,  f  n  is  any  positive 
integer  ; 

from  (5)  X  -\-y  is  a  factor  of  x"  —  y",  if  n  is  even  ; 

fro7n  (G)  X  +y  is  a  factor  of  x"  +y",  if  n  is  odd. 

Comparing  with  the  type  forms  of  p.  101,  we  see  that 
x^  —  y^  falls  under  (2)  and  under  (5)  ;  in  fact  x^  —  //^  is 
divisible  by  x  —  y  and  by  x-\-y,  as  we  already  know.     The 


332  PROGRESSIONS   OR  SE(iUENCES         [Ch.  XIII 

form  x'^  —  i/^  fulls  under  (2)  only  ;   it  is  divisible  by  x  —  y^ 
and  the  quotient  \'&  7? -\-  xy  ■\-  y"^. 

Likewise, 

x^  +  y^  =  {x  +  y^Qc^  -  ^y  +  f-)  %  (6)  (see  p.  101); 

x^  _  /  =  (a:  -  y)(^x^  +  x^y  +  a:/  +  y^)  by  (2), 

or  also,        =  (a;  +  y^{^^  —  x'^y  +  xy^  +  i/^)  by  (5), 

or  also,        ={x-  y)(^x  +  y){x'^  +  y''-)  by  (2)  and  (5) , 

^-y^=ix  —  y^ (2;4  +  a%  +  i%2  ^  ^-j/S  +  t/4) ; 

x^  ^-y^={x-\-  y)  (3;4  -  a;3y  +  x^y'^  —  xy^  +  y*), 

and  so  on.     Compare  also  Appendix,  §  6. 

Ex.  1.    Factor  32  a^  -  Z^^ 

This  compares  to  (2)  ii  x  =  2a,  y  =h,  /i  =  5  ;  hence, 
32  a^-b^  =  (2  a  -  h)  [(2  «)*  +  (2  aYh  +  (2  fO^//-^  +  (2  u)h^  +  h^-} 
=  (2  a  -  6)(16  a*  +  8  a^i  +  4  a%-^  +  2  ah^  +  ^/^). 

Ex.  2.    Factor  81 1*  -  s«. 

This  corresponds  to  (2)  if  x  =  3  ^  ^  =  s^,  n  =  4 ;  hence,  x  —  y=Zt  —  s^ 
is  a  factor.  By  (5)  x  +  i/  =  3  <  -|-  s^  is  also  a  factor ;  hence,  (3  t  —  s-) 
(3 1  +  s'^)  is  also  a  factor,  i.e.  9  ^^^  —  s*  is  a  factor.  This  results  also  by 
taking  x  =  9fi,  y  =  s*,  n  =  2,  so  that  .r^  —  ?/2  =  81 1*  —  s^  From  either 
we  find  81 1*  - .«»  =  (9  /^  _  s4)(9  fi  +  s")  =  (3  /  -  s2)(3  t  +  s'-i)(9  /'^  +  *■"). 

EXERCISES  III:  CHAPTER  XIII 

Factor  the  following  expressions  : 

1.  x^-f.  5.  a;"  ±.7".  9.  1  +  1024  3^. 

2.  1+1/^.  6.  8  7>r+27  7i^  10.  l-G4xV. 

3.  x^-2/'.  7.  81a*- 16  6^  11.  32 /u^  +  243  u^"^ 

4.  l-.y^l  8.  12oa'7/'-l.  12.  G2r>.ry-r« 


101]  SUMMARY  333 


SUMMARY  OF  CHAPTER  XIII:  SEQUENCES  OR  PROGRES- 
SIONS, pp.  323-;5:V2 

Sequence  (or  Prn/jresslon)  :  set  of  numbers  in  definite  order. 

§  155,  p.  328. 

Arithmetic  Sequence:  common  difference  between  terms. 

§  156,  pp.  323-324. 

Formukc  :  t„  =  a  +  (n  -  l)r/;  s  =  l"](n -{- 1).     Exercises  I. 

^-^  §  157-158,  pp.  321-327. 

Geometric  Sequence  :  constant  common  ratio.  §  159,  p.  328. 

Formulre  :  t"  =  ar"  '^ ;    /  =  ar"  ^ ;  .s  =  -i -^  •     Exercises  II. 

(^  ~  '■)     §  160,  pp.  328-331. 
Factoring  jc'"  ±  i/":  z  —  //  a  factor  of  x"  —  ?/"  ;  x  +  y  a  factor  of  x"  — »/" 
•when  n  is  even  ;  of  x"  +  ^"  when  n  is  odd.    Exercises  III. 

§  161,  pp.  331-332. 


CHAPTER   XIV 

LOGARITHMS 

162.  Introduction.  We  shall  now  study  a  method  by 
which  long  numerical  computations  are  greatly  simpli- 
fied. Before  proceeding  with  this  chapter,  the  student 
should  review  the  entire  subject  of  exponents.  (See  §§ 
135-144.)  The  following  examples  may  serve  as  a  basis 
of  this  review. 

EXERCISES    I:     CHAPTER    XIV 

1.  SimpUfy  the  following  as  much  as  possible  without  per- 
forming any  long  numerical  computations  : 

a.  x'y^x\    b.  127^x1271     c.  74' x  74.    d.  3-'x2-"';  7»  X  5». 
e.  3^'  X  3"^.     Can  this  be  done  in  two  ways  ? 
/.  g'-^cf.        g-  (a"')'.       h.  -\/a^-.         i.  Va''.      ./•  x^^x^. 
Ic.  24»  --  6» ;  243^  H-  81«.     I.  127''  h-  127^     Can  this  be  done  in 
two  ways  ? 

m.   v'S^^  n.  -v/27''.     Can  this  be  done  in  two  ways  ? 

2.  Express  without  an  exponent: 

a.  S-'.       b.  8i       c.  8"'.       (1.  125".       e.  (j\y.      f.  (^fK 

3.  Sum  up  all  the  principles  you  have  used  in  Exs.  1  and 
2  by  completing  the  following  ecpuations.  Tell  which  of  the 
above  examples  are  special  cases  of  each : 

a.  x-"'=  c.  .x"=  e.  .x-'"-r-.T"=  g.  a/.i;"'  = 

m 

b.  X''  =  (I  of  •  ;r"  =         /.   (.i;")"  =  h.  X-"  = 

4.  State  in  words  the  laws  expressed  in  Ex.  3.  (See  §§  135, 
136,  143.) 

3.34 


uh;  AH  ITIIMS  335 

5.    (■()iiii)lete  tlie  followiiii,'  t;il)le  by  giving  all  the  powers  of 
1(3  at  intervals  of  \  from  —  4  to  +  4. 

k;"  =  1  ir,"  =  1 

16?  =  4  1G"2  =  1. 

16^  =  8  16-t  =  | 


16*  =  65536  16-"  = 


6  5536 


6.  By  means  of  the  table  in  example  5  find  the  value, 
wherever  it  can  be  done  without  any  long  computation,  of 
the  following  expressions : 

a.  512  X  128  =  what  ? 

Solution      512  x  128  =  10?  x  16*  =  16^  +  *  =  Ifi*  =65536. 

b.  16384-64.  d.  S/128  x  16384»- 
^   512x32768.  ,,  ^(W.^rx(8)". 

8192  v'^uao/  ^   / 


/ 


16384 
2048 


-^  X  8192 
2633 


The  preceding  metliod  alone  is  not,  completely  successful  in  this 
case,  because  we  cannot  perform  the  subtraction  In'  means  of  expo- 
nents. 

g.  Can  the  following  operations  be  performed  by  means  of 
your  table  ?     Wliy  not  ?     ^1024,  a/572. 

163.  A  Table  of  Exponents.  Our  table  is  not  suffi- 
ciently complete  for  practical  purposes,  as  was  seen  when 
we  attempted  to  solve  6  g,  above,  by  means  of  it.  In  order 
to  make  a  table  that  Avill  be  of  practical  use  in  examples 
involving  any  number,  it  will  be  more  convenient  to  use 
10  as  a  basis  instead  of  16,  as  10  is  better  suited  to  the 

\ 


336 


LOGARITHMS 


[Cii.  XIV 


decimal  system.     Let  the  student  complete  the  following 
table  and  make  himself  familiar  with  it: 


10^=1 

101=10 

102=100 


10*^    =1 
10-1=. 1 
10-2  =.01 


106  ^  1000000 


10-6  =.000001 


We  may  use  this  table  as  we  did  the  other  in  solving 
certain  simple  problems,  but  it  has  the  same  difficulty  as 
the  other,  for  such  numbers  as  273,  1772,  etc.,  are  not 
found  in  it ;  but  it  has  an  advantage  over  the  other  table 
in  that  it  is  easier  to  remember. 

Let  us  now  make  a  similar  table  containing  many  more  numbers. 
To  do  this,  let  us  make  a  graph  of  the  table,  in  other  words,  a 
graph  of  the  equation  —  aol 

The  preceding  table  gives  values  of  n  and  of  L  as  follows : 


■  Point  in  Fiij;.  (58  : 

M 

L 

K 

A 
0 

B 

1 

C 
2 

L: 

-6 

—  •J 

-4 

-3 

-2 

-1 

3 

4 

5 

6 

n  : 

.0000001 

.1 

1 

10 

100 

1000000 

[Let  the  student  complete  the  table.] 

Since  n  can  never  become  negative,  we  take  the  starting  point  near 
the  left  edge  of  the  jiaper. 


1 

— 

T 

~ 

_ 

~ 

■" 

~ 

~ 

^ 

■" 

"' 

■ 

■ 

n 

ti 

- 

- 

_ 

^ 

_ 

— 

■^ 

F-" 

— 

L. 

^ 

^ 

■" 

■ 

- 

^ 

— 

-^ 

■ 

m 

6 

r 

* 

^ 

It 

r 

O 

. 

. 

E; 

* 

( 

) 

10 

t 

^ 

r 

1 

/ 

K 

J 

_ 

_^ 

_j 

_ 

__ 

Fig.  68. 


1G3] 


LOGARITHMS 


337 


There  is  difficulty  in  making  a  graph  of  the  whole  table,  owing 
to  the  large  numbers  that  occur;  moreover,  we  have  not  enough 
points  to  draw  the  curve  very  accurately ;  but  the  student  will  be  able 
to  answer  the  following  questions  : 

How  does  n  change  as  L  becomes  very  large  positively?  How  high 
does  the  curve  rise?  How  does  n  change  as  L  becomes  very  large 
negatively?  Does  it  ever  touch  the  vertical  main  line?  How  close 
does  it  get  to  it?  How  large  a  sheet  of  paper  would  be  needed  to  plot 
all  the  points  of  the  table?     How  finely  would  it  need  to  be  ruled? 

Now  let  us  draw  the  part  of  the  curve  from  A  to  B  carefully. 
Since  the  curve  is  so  flat,  we  shall  keep  the  same  scale  on  the  horizon- 
tal line,  and  take  the  scale  ten  times  as  great  on  the  vertical  line ;  and 
we  shall  take  the  starting  point  in  the  lower  left-hand  corner  of  our 
paper.  We  will  try  fractional  values  of  L.  Let  L  —  .5  —  |.  Then 
n  =  \0i  =:  VTO  =  3.16.  Let  L  =  .25  =  \.  Then  n  =  10?  ^  (10^)^  = 
(;5.16)2  =  VIUU  =  1.78+.  Let  L  =  .75  =  ^.  Then  n  =  10?  =  10^  x  loi 
=  3.16  X  1.78  =  5.62.  Continuing  this  process,  we  find  the  values  of 
n  and  L  as  given,  in  the  following  table,  the  computations  being  made 
in  the  order  indicated  by  the  letters  of  the  alphabet. 


Point  in  Fig.  69  : 

A 

F 

D 

G 

C 

H 

E 

/ 

B 

L  (fractionally): 

0 

i 

i 

1 

i 

f 

3 

? 

i 

1 

L  (decimally): 

0.000 

.125 

.250 

.375 

.500 

.625 

.750 

.875 

1.000 

*/  (to  two  places): 

1.00 

1.33 

1.78 

2.37 

3.16 

4.22 

5.62 

7.50 

10.0) 

The  points  /,  K,  L,  M,  N,  0,  P,  Q,  in  the  figure  correspond  to 
values  of  L  by  sixteenths;  the  corresponding  values  of  n  may  be  com- 
puted by  the  student,  or  read  off  from  the  figure.  The  student  should 
plot  these  values  with  great  care,  draw  a  smooth  curve  through  them, 

I    and  keep  the  figure  for  his  own  use  in  what  follows.     It  should  be 

:    like  Fig.  69. 

j  We  can  now  determine  by  measurement  from  the  fig- 
;  are  the  approximate  value  of  the  exponent  L  when  the 
'  number  n  is  sfiven  between  1  and  10  ;  and  of  n  when  L  is 
!  given  between  0  and  1.  The  exponent  L  is  often  called 
the  logarithm  of  w,  and  we  write  L=  log  n.      (See  §  10  i.) 


338 


LOGARITHMS 


[Cn.  XIY 


tcij/r 


-1^0 


w. 


-M» 


fe: 


K 


r'Q 


m 


'H 


Fig.  (i9. 


Ex.  1.    Find/,  if  10^  =  4. 

"We  measure  4  to  the  right  on  the  main  horizontal  line,  and  then 
the  distance  up  to  the  curve  is  the  required  value,  L  -  .60  (nearly). 
Hence,  .60  is  the  logarithm  of  4  (nearly),  or  log  4  =  .60  (nearly). 

Ex.  2.    Find  n  =  lO'K 

AVe  measure  .53  up  on  the  main  vertical  line,  and  then  the  dis- 
tance on  a  horizontal  line  to  the  curve  is  tlie  rec^uired  number, 
n  =  3.4  (nearly).     Tliat  is,  .53  =  log  3.4  (nearly). 

Contiiiuinf^  in  a  systematic  manner,  by  actual  measui-e- 
nient  on  the  fignre,  as  above,  we  find  log  1  =  0;  log  1.1  = 
.04;  log  1.2  =.08;  log  1.3  =.11. 


163] 


hOGAKITlliMS 


3^9 


These  values  can  be  conveniently  arranged  in  tabular 
form,  called  a  table  of  exponents,  or  a  table  of  logarithms. 
In  the  table  which  follows,  the  values  are  correct  to  two 
places  of  decimals. 

Two  Place  Table 


71 

0 

.1 

.2 

.3 

.4 

.5 

.6 

.7 

.8 

.9 

1 

0 

.01 

.08 

.11 

.15 

.18 

.20 

.23 

.26 

.29 

2 

.30 

..32 

.34 

.36 

.88 

.40 

.43 

.45 

.46 

3 

.48 

.49 

.51 

.52 

.53 

.56 

.,57 

.58 

.59 

4 

.60 

.61 

.62 

.63 

.64 

.66 

.67 

.68 

.69 

5 

.70 

.71 

.72 

.73 

.74 

.75 

.76 

.77 

6 

.78 

.79 

.80 

.81 

.82 

.83 

.84 

7 

.85 

.86 

.86 

.87 

.88 

.89 

.90 

8 

.90 

.91 

.92 

.93 

.94 

.95 

9 

.96 

.97 

.97 

.98 

.98 

.99 

10 

1 

[Let  the  student  fill  in  the  blank  spaces  from  his  figure.  Notice 
that  the  value  to  be  inserted  must  lie  between  those  on  either  side  of 
it.  The  values  obtained  by  the  student  from  his  figure  may  not  agree 
exactly  with  those  given  here,  on  account  of  the  inaccuracy  of 
drawing.] 

To  find  from  the  table  log  6.5  we  look  for  the  6  in  the 
column  marked  w  ;  opp'osite  6  and  in  the  column  marked 
.5  we  find  .81,  wliich  is  the  logarithm  of  (5.5. 

Values  not  actually  in  the  table  may  also  be  found  by  a 
process  of  reasoning  explained  in  the  following  examples: 


Ex.  3.    Find  »  =  10-' 

„  ^  102-53  =    10-^   X    10-53 


100  X  3.39  =  .339,  nearly;  that  is,  2..53 


log  .339. 


340  LOGARITHMS  [Ch.  XIV 

Ex.  4.   Find  L  if  10^  =  7250. 

We  find  from  the  table  that  7.25  =  10-^,  nearly.  But  1000  =  lO^. 
Hence, 7250  =  1000  X  7.25=  103  xlO»5=  103-86.  Hence,  Z  =  3.S6.  That 
is,  log  7250  =  3.86. 

Ex.  5.   Find  log  (.0725);  that  is,  find  L  if  10^  =  .0725. 

As  above,  7.25  =  lO-s^,  .01  =  IO-2.  Hence,  .0725  =  .01  x  7.25  = 
10-2  X  10-8«  =  10-2+-86.  Hence,  log  .0725  =  -  2  +  .86.  It  is  usually 
more  convenient,  when  the  integral  part  of  the  logarithm  is  negative, 
to  leave  the  decimal  part  always  positive,  instead  of  adding ;  thus,  we 
write  -  2  +  .86  and  not  -  1.14. 

164.    Simple  Computation  by  Exponents.     We  can  now 

perform  any  of  the  simple  operations  of  multiplication, 
division,  raising  to  powers,  or  extraction  of  roots  upon 
any  number  whatever  very  easily  by  logarithms,  though 
our  results  will  not  be  precisely  accurate. 

Ex.  1.   Multiply  2  by  3. 

We  can  do  this  easily  without  logarithms:  2x3  =  6.  Notice 
that  2  =  103«,  3  =  10-^8.  Hence,  2x3  =  10-9>  x  lO--^  =  lO-^s  =  6.  Let 
the  student  make  the  measurements  on  liis  figure.  The  student's 
measurements  may  not  agree  precisely  with  these,  owing  to  inaccuracy 
in  the  figure  used. 

Ex.  2.   Multiply  3|  by  7.3. 

Let  the  student  use  his  own  figure.  3|  =  lO-^e,  7.3  =  10-86.  Hence, 
3|  x  7.3  =  101*2  =  l()i  X  10*2  =  10  x  2.6  =  26.  Multiplying  by  the 
ordinary  method,  we  see  that  this  is  not  precisely  accurate,  owing  to 
the  inaccuracy  of  our  table,  but  this  degree  of  accuracy  would  be 
sufficient  in  many  practical  problems. 

Ex.  3.    Divide  13  by  7. 

13  =  10  X  1.3  =  101  X  i().u  =  iQUi;  7  =  10-85.     Hence,  13  ^  7  =  lOi" 

■-  10-85=  10-26=  1.8^  nearly. 

Ex.  4.    Find  (4.3)". 

4.3  =  106-\  Hence,  4.-3"  =  lOi^^-^s  =  lO""  =  lO^  x  10 ^^  = 
10000000000  X  5.1  =  51000000000.     Of  course,  this  is  far  from  accu- 


lG;i-lU5]  LOGARITHMS  341 

rate  ;  we  are  sure  only  of  the  first  one  or  two  figures,  but  tlie  student 
should  notice  tlie  great  saving  in  labor,  and  that  frequently  the 
accuracy  here  attained  would  be  sufficient.  With  the  more  extended 
table  given  below,  greater  accuracy  is  obtainable. 

Ex.  5.    Extract  the  seventh  root  of  7825. 

The  work  would  be  extremely  long  by  methods  previously  used. 
We  have  here 

7S25  =  1000  X  7.825  =  10^  x  lO-s^  =  lO^**,  nearly. 

Hence,  -v/7825  =  VlO-''**  =  10*'  =  3.6,  nearly.  In  finding  the  loga- 
rithm of  7.825  by  the  table  we  took  the  nearest  logarithm  found  in 
the  table.     A  more  accurate  table  will  be  found  on  p.  348. 


EXERCISES    II:    CHAPTER   XIV 

Simplify  the  following  by  the  aid  of  the  preceding  table  ; 
compare  each  result  with  the  figure : 


1.    4.3x2.3.  4.    V1730.  7.   45-+ V (1.5)-'. 

Z.    230  +  17.  5.    2.3'.  8.    (3-^5)  • 

3.   2.7^-.  6-    V2^.  9.    .005^x5200^ 

^Q     -e/31  X  4.1  (4J)i, 

165.  Definitions  and  Principles.  A  fundamental  num- 
ber must  always  be  chosen  as  the  base;   this  is  tisually  10. 

Tlie  logarithm  of  a  number  to  a  given  base  is  the  exponent 
with  ivhicli  the  base  must  be  affected  to  produce  the  number. 

In  other  words  10^  =  n  and  log  n  =  L  have  precisely  the 
same  meaning.     We  have  taken  10  as  our  base,  as  usual. 

Henceforth,  when  the  base  is  not  specified,  it  will  be 
understood  that  the  base  10  is  used.      AVHieu  the  base  10 


342  LOGARITHMS  [Cn.  XIV 

is  used  the  logarithms  are  called  common  logarithms  or 
Briggs's  logarithms. 

Note  1.  The  base  10  is  most  common,  but  others  may  be  used. 
When  necessary,  we  write  logj^/t  to  avoid  ambiguity.  In  general,  if 
the  base  is  /;,  tlie  logarithms  are  denoted  by  log/,  n.  The  base  nuiy  be 
any  number  except  0  or  1. 

Note  2.  We  have  not  explicitly  defined  the  meaning  of  an  irra- 
tional exponent,  and  the  student  should  not  be  burdened  at  this  stage 
with  the  idea.  It  is  sufficient  to  call  attention  to  the  fact  that  the 
use  of  the  smooth  curve  (§  163)  through  certain  points  that  can  be 
located  involves  the  essential  idea  of  a  rigorous  treatment.  (See 
Appendix,  §  30.) 

From  the  principles  of  exponents  (§§  104, 135,  pp.  193, 
285,  and  p.  334),  it  is  evident  that 

I.    The  logarithm  of  a  product  is  equal  to  the  sum  of  the  . 
logarithms  of  the  factors,  for  10'"  x  10"  =  10'"+".  ^ 

11.  The  logarithm  of  a  quotient  is  equal  to  the  logarithm 
of  the  dividend  less  that  of  the  divisor,  for  10"'  h-  10"  =  10'""". 
III.  The  logarithm  of  a  power  of  a  number  is  equal  to  the 
logarithm  of  the  number  multiplied  by  the  exponent  of  the 
power,  for  (10"*)"  =  10"^".  This  principle  includes  the 
extraction  of  roots  by  using  fractional  exponents,  for, 
by  §  139, 

1  „1  7» 

—  m  X  -  — 

(10™)«  =  10     "  =  10". 

These  may  be  stated  as  follows : 
I.    Log  (m  X  n)  =  log  m  +  log  n. 

IT.    Log      =  log  m  —  log /J. 
n 

III.    "Log  n'^  =  K  log  n .  (AT  may  be  an  integer  or  a  fraction.) 

The  logarithm  of  1  to  any  base  is  zero.     Since  Ifi  z=\. 
The  logarithm  of  the  base  is  1.  Since  ¥  =  b. 

The  common  logarithm  of  a  number  between  one  and  ten 
lies  between  zero  and  one.     See  the  table  or  the  iigure. 


IGfj-lGG]  LO(iAin'l'll.M.S  343 

166.  Characteristic  and  Mantissa.  Let  us  now  study 
the  curve  lU^=  n,  or  log  n  =  I  for  a  greater  range  of  values. 
From  the  table  already  constructed  (§  163,  p.  339)  we  find, 
for  example,  10-'*^  =  3.4.     Hence, 

101-53  =  34.  10-i+-53=.34 

10253=340.  10-2+ 53  =034 

103-53  =  3400.  10-3+.53  =  ,  0034 

104-53  =  .34000.  10-4+53  =  .00034 


The  student  should  notice  that  increasing  the  logarithm 
by  1  corresponds  to  multiplying  the  number  by  ten  (i.e.  the 
decimal  point  is  changed  one  place).  Likewise,  decreasing 
the  logarithm  by  1  corresponds  to  dividing  the  number  by 
ten. 

If  l<x<  10,  log  a:  =  0  +  a  positive  fraction. 
If  10<.?<  100,  log  2;=  1  +  a  positive  fraction. 
If  100  <  X-  <  1000,  log  x=      2  +  a  positive  fraction. 

If  .1  < a: <  1,  log  x=  —  1+  a  positive  fraction. 
If  .01<2;<       .1,  log  a:  =  —  2  4-  a  positive  fraction. 

The  fractional  part  will  be  the  same  in  all  these  cases  if 
the  digits  in  the  timnber,  x,  are  the  same  and  follow  each 
other  in  the  same  order.  The  position  of  the  decimal  point 
determines  only  the  integral  part  of  the  logarithm. 

The  fractional  part  is  called  the  mantissa,  and  is  gen- 
erally taken  positive  to  avoid  the  introduction  of  a  differ- 
ent decimal ;  it  is  determined  by  the  digits  in  the  number 
as  given  and  does  not  depend  on  the  position  of  the  deci- 
mal point.  A  table  is  needed  to  find  the  value  quickly. 
I  The  integral  part  is  called  the  characteristic ;  it  may  be 
j  either  positive  or  negative.     This  integral  part  can  always 


344 


LOGARITHMS 


[Ch.  XIV 


be  found  by  inspection,  as  above ;  hence,  a  table  of  loga- 
rithms contains  only  the  fractional  parts  of  the  logarithms. 
Let  us  now  change  our  scale,  using  1  =  5  small  spaces  on 
the  vertical  line  and  1  =  1  small  space  on  the  horizontal  line. 
Plot  a  sufficient  number  of  points  by  means  of  the  table  on 
p.  839  and  the  principle  just  given.  Then  on  the  same 
sheet  plot  the  characteristic.  This  gives  the  stair  step 
bounding  the  shaded  region.  The  distance  from  the  main 
horizontal  line  up  or  doivn  to  the  stair  step  is  the  charac- 
teristic, the  distance  from  the  stair  step  up  to  the  curve  is 
the  mantissa. 

In  making  a  table  of  logarithms  it  is  customary  to  give 
only  the  mantissa,  omitting  the  decimal  point  both  in  the 

number  and  in 
the  logarithm, 
as  mentioned 
above.  The 
student  is  left 
to  determine 
the  character- 
istic. This  is 
the  same  as 
giving  the  logarithms  of  numbers  between  one  and  ten. 

167.  Four-place  Table  of  Logarithms.  The  table  on 
pp.  348-349  is  constructed  on  the  same  plan  as  that  on 
p.  339.  The  logarithms  are  given  to  four  places  of  deci- 
mals and  the  corresponding  numbers  are  given  to  three 
figures.  If  greater  accuracy  is  required,  tables  can  be 
bought  to  five,  six,  seven,  or  ten  places.  The  logarithms 
as  given  in  the  following  table  cannot  generally  be  exact, 
as  all  figures  after  the  fourth  place  are  rejected.  If  the 
fifth  figure  is  5  or  more,  the  fourth  figure  is  increased  by  one. 

Thus,  if  the  logai'ithm  of  a  number  is  .437826  •••,  it  is  given  in  the 
table  as  .4378    If  the  logarithm  is  .43455  •••,  it  is  given  as  .4346 


Fig.  70. 


1G6-167] 


LOGARITHMS 


345 


To  find  the  Logarithm  of  a  Given  Number  from  the  Table. 

Ex.  1.    Find  the  logarithm  of  3. 

In  the  columns  marked  N  we  look  for  30  (the  decimal  point  after 
the  3  is  omitted).  On  a  line  with  this  in  the  column  headed  0  we 
find  4771.     Hence,  log  3  =  .4771 

Ex.  2.   Find  the  logarithm  of  4.6 

In  the  column  marked  N  we  find  46.  On  a  line  with  it  in  the 
column  marked  0  we  find  6628.     Hence,  log  4.6  =  .6628 

Ex.  3.    Find  the  logarithm  of  3.76 

In  the  column  marked  A'' we  find  37.  On  a  line  with  this  in  the 
column  marked  6  we  find  5752.     Hence,  log  3.76  =  .5752 

Ex.  4,    Find  the  logarithm  of  3760. 


By  example  3, 10-^^52  ^  3.75 
But,  103=1000 


.} 


Hence,  103-5752  ^  3760 
or,  W  3760  =  3.5752 


Ex.  5,    Find  the  logarithm  of  .0376. 

By  example  3,   10-5"52  =  3.7fj  l      Hence,  10-2+-5T52  ^  .0376 

But,  10-2  ^  .01    J  or,  log  .0376  =  -  2  +  .5752 

This  is  sometimes  written  los;  .0376  =  8.5752  —  10. 


Ex.  6.  Find  log  3764, 
From  the  table 
log  3.76  =  .5752  (.4) 
log  3.77  =  .5763  (B) 
We    will    plot   these 
points  on  a  very  large 
scale,    drawing    only    a 
small  i^art  of  the  curve. 
A  is  the  point  where 
n  =  3.76,    L  =  .5752 
B  is  the  point  where 
n  =  3.77,    L  =  .5763 
E  is  the  point  where 

n  =  3.764 
"We  wish  to  find  L. 


~~ 

~ 

^ 

~ 

~ 

~ 

— 

~ 

~ 

~ 

~ 

~ 

L 

Ic 

K 

n 

H 

/ 

^ 

/ 

.57(30 

..i 

f- 

^ 

CI 

/^ 

A 

0 

P 

J\ 

'of 

3 

^ 

■T 

-i 

.57.>5 

/ 

vr 

'/ 

/ 

/ 

^, 

/ 

D 

, 

T 

<- 

,004 

\( 

'^ 

= 

.57o 

0 

(—, 

n 

Oi 

_ 

_ 

1) 

V 

3. 

75 

' 

~ 

3^ 

75 

V 

31 

7G 

~ 

i 

(76 

5 

~ 

~ 

F7 

7 

— 

cale 

01 

b< 

>n 

F° 

Qtt 

lUv 

, 

' 

'' 

|.U01 

y 

"■li^a 

ly 

_^ 

. 

1    1 

Fig.  71. 


34H  LOGARITHMS  [Ch.  XIV 

If  we  draw  the  straight  line  AB,  cutthig  DE  at  F,  then  F  will  lie 
very  near  to  E  and  it  will  be  sufficiently  accurate  to  find  the  heiglit 
of  F  instead  of  E.  D  is  just  as  high  as  A,  i.e.  .5752;  CB  =  .0011  ; 
DF  —  .4  of  .0011  =  .0004  (dropping  decimal  places  after  the  fourth). 

Hence,  the  height  of  F  is  .5752 +  .0004  =  .5756 
That  is,  10-6™  =  3.764 

103  =  1000. 
Hence,  lO^-^^se  ^  3764. 

Hence,  log  3764  =  3.5756 

This  process  of  finding  a  result  not  given  explicitly  in  the  table  is 
called  interpolation. 

Ex.  7.    Find  the  logarithm  of  .02756 

Consider  2.756  log  2.75  =  .4393 

log  2.76  =  .4409 

.0016 

log  2.756  =  .4393  +  (.6  x  .0016)=  .4393  +  .0010  (nearly)  =  .4403 

10-4*03  =  2.756     Hence,  10-2+-"03  ^  .02756 

Hence,  log  .02756  =  -  2 +  .4403  =  8.4403  -  10.  The  student  should 
draw  a  figure  illustrating  this  work. 

To  find  the  Number  corresponding  to  a  given  Logarithm. 

When  the  decimal  part  of  the  logarithm  can  be  found 
in  the  table,  the  corresponding  number  can  be  written 
down  at  once. 

Ex.  1.   Find  n  if  log  n  =  4.9175 

Looking  in  the  table,  we  find  9175  on  a  line  with  82  (in  column  iV) 
and  in  the  column  marked  7.  Hence,  lO-^i'^  =  8.27  Hence,  10*-9i"  =: 
82,700.  Hence,  log  82,700  =  4.9175  When  the  decimal  part  of  the 
logarithm  cannot  be  found  in  the  table,  we  follow  a  process  similar 
to  that  in  examples  6  and  7  above. 

Ex.  2.   Find  n  if  log  n  =  2.4574 

From  the  table  we  find  log  2.87  =  .4579  log  n  =  2.4574 

log  2.86  =  .4564  log  286  =  2.4564 

Difference  =  .0015  =    .0010 


167-168]  LOGARITHMS  347 

The  difference  between  log  286  and  log  n  =  .0010  This  is  f§ 
of  the  difference  between  log  286  and  log  287.  Hence,  .4574  =  log 
(2.86  +  ig  of  .01).  Hence,  2.4574  =  log  286.7  The  student  should 
construct  a  figure  like  that  above  for  this  case. 

Xotice  that  the  differences  found  above  are  given  in  the  column 
marked  D  in  the  table.  These  differences  are  not  always  exact,  since 
the  real  differences  often  vary  in  the  same  row ;  but  they  are  always 
sufficiently  accurate  for  use  with  this  table. 

EXERCISES   III:    CHAPTER  XIV 

1.  Find  the  logarithm  of : 

(a)  7  (c)  600  (e)    .032  (g)  .0467  (i)  2.473 

(6)  40         (d)  4.7  (/)  2.43  (h)  57,200        (j)  .04257 

2.  Find  the  number  Avhose  logarithm  is  : 

(a)  .3010        (c)   2.3802      (e)   5.4533      (cj)  2.6290       (i)  2.0027 
{b)  3.4771      (d)  3.3075      (/)  1.5732      (/<)  1.4563       (J)    .0317 

168.    Computation  by  Logarithms. 

Ex.  1.    Simplify  ^2471x3.428. 
^     ^    V      (16.3)2 

'/247  X  :j.428_^i'T0^9--"  x  lO-saso       3;jq2.9277 

^         (16.8)2  "-\  (^lQl.-2V22y  -    \  102.4244 

=   ^10-5033  :^  10-16""  =  1.471 

This  work  may  be  tabulated  as  follows  : 


log  Y 


log  247  =  2..3927 
log  3.    428  =  .5350 


Divide  by  3 : 


log  numerator  =  2.9277 

log  16.3  =  1.2122 
log  (16.3)2  = 
log  quotient  = 


log  answer  =  .1678 

hence,  answer  =  1.471 


348 


LOGARITHMS 


[Ch.  XIV 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

42 

11 

0414 

0453 

0492 

0531 

0569 

0(J07 

0645 

0682 

0719 

0755 

38 

12 

0792 

0828 

0864 

0899 

09:34 

09(i9 

1004 

1038 

1072 

1106 

35 

13 

li:5y 

1173 

1206 

1239 

1271 

1303 

13:55 

131)7 

1:599 

14:50 

32 

14 

14tJl 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

30 

15 

17G1 

1790 

1818 

1847 

1875 

um 

1931 

1959 

1987 

2014 

28 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

26 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

25 

18 

2o53 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

24 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

22 

20 

3010 

3032 

3054 

3075 

3096 

3118 

31:39 

3160 

3181 

3201 

21 

21 

3222 

3243 

3263 

3284 

3:304 

3324 

3:345 

3:365 

3385 

3404 

20 

22 

3424 

3444 

3464 

3483 

:«02 

3522 

3541 

35(50 

3579 

3598 

19 

23 

3617 

3(j36 

3(  155 

3674 

3692 

3711 

;3729 

3747 

3766 

3784 

18 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3<t09 

3927 

3945 

3962 

18 

25 

3979 

3997 

4014 

4031 

4048 

40(35 

4082 

4099 

4116 

41:33 

17 

26 

4150 

4166 

4183 

4200 

4216 

42:32 

4249 

4265 

4281 

4298 

16 

27 

4314 

4330 

4346 

4362 

4:378 

4:393 

4409 

4425 

4440 

445(5 

16 

28 

4472 

4487 

4502 

4518 

4533 

4548 

45(i4 

4579 

4594 

4(509 

15 

29 

4(324 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

15 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

14 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

50:58 

14 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5i:i2 

5145 

5159 

5172 

13 

33 

5185 

5198 

5211 

5224 

52:37 

5250 

5263 

5276 

5289 

5302 

13 

34 

5315 

5328 

5340 

5353 

5366 

5:378 

5391 

5403 

5416 

5428 

13 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

12 

36 

5563 

5575 

5587 

5599 

5611 

562:5 

56:55 

5647 

5658 

5670 

12 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

57(33 

5775 

5786 

12 

38 

5798 

5809 

5821 

6832 

5843 

5855 

5866 

5877 

5888 

5899 

11 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

11 

40 

6021 

G031 

6042 

6053 

6064 

(i075 

6085 

609(5 

6107 

6117 

11 

41 

6128 

6138 

6149 

6160 

6170 

(il80 

6191 

6201 

6212 

6222 

10 

42 

6232 

(5243 

6253 

6263 

6274 

6284 

()294 

(3:;o4 

6314 

6:525 

10 

43 

6335 

6345 

6355 

6365 

6375 

6:385 

(i:;95 

6405 

6415 

6425 

10 

44 

6435 

6444 

6454 

6464 

6474 

(i484 

6493 

(5503 

6513 

6522 

10 

45 

6532 

6542 

6551 

6561 

6571 

(3580 

6590 

6599 

6609 

6618 

10 

46 

6628 

6637 

6646 

(i(556 

()()(i5 

(3(375 

6684 

6693 

(5702 

6712 

9 

47 

6721 

(5730 

(i739 

()749 

6758 

67(37 

()776 

6785 

(5794 

(5803 

9 

48 

6H 12 

6821 

6830 

()839 

6848 

6857 

68()6 

6875 

6884 

(5893 

9 

49 

(;'.K)2 

()911 

6920 

6928 

6937 

6946 

6955 

6SH54 

6972 

6981 

9 

50 

6990 

G998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

9 

51 

707f) 

70H4 

7093 

7101 

7110 

7118 

712(; 

71:55 

714:5 

7152 

8 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

722(5 

7235 

8 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7:500 

7:508 

731() 

8 

54^ 

7324 

7;i32 

7;}40 

7348 

735(i 

7:564 

7372 

7:580 

7388 

73% 

8 

16S] 


KOIR    PLACE   TABLE 


849 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D 

55 

7404 

7412 

7419 

7427 

74:55 

7443 

7451 

7459 

7466 

7474 

8 

56 

7482 

74iK) 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

8 

57 

7559 

75<i() 

7574 

7582 

7589 

7597 

7(504 

7612 

7619 

7627 

H 

58 

7(534 

7(J42 

7649 

7(557 

7(5(54 

7672 

7(579 

768(5 

7694 

7701 

7 

59 

7709 

771(5 

7723 

7731 

77:^8 

7745 

7752 

77(50 

7767 

7774 

7 

60 

7782 

7789 

779(5 

7803 

7810 

7818 

7825 

7832 

78:39 

7846 

7 

61 

7853 

78(50 

7868 

7875 

7882 

7889 

78<t(5 

7903 

7910 

7917 

7 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7«t()6 

7973 

7980 

7987 

7 

63 

7CKJ3 

8000 

8007 

8014 

8021 

8028 

80:55 

8041 

8048 

8055 

7 

64 

8062 

80(59 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

7 

05 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

7 

66 

8195 

8202 

8209 

8215 

8222 

8228 

82:55 

8241 

8248 

8254 

7 

67 

82(31 

82(57 

8274 

8280 

82.S7 

8293 

8299 

8:306 

8312 

8319 

6 

68 

8325 

8331 

8338 

8:V44 

8:351 

8:357 

8;;63 

8370 

8376 

8382 

6 

69 

8388 

8395 

8401 

8407 

8414 

8420 

842(5 

8432 

8439 

8445 

6 

70 

8451 

8457 

84(53 

8470 

8476 

8482 

8488 

84f)4 

8500 

8506 

6 

71 

8513 

8519 

8525 

8531 

85:57 

8543 

8549 

8555 

8561 

8567 

6 

72 

8573 

8579 

8585 

8591 

8597 

8(503 

8(509 

8615 

8621 

8627 

6 

73 

8(533 

8639 

8645 

8651 

8657 

86(53 

8(569 

8675 

8681 

8686 

6 

74 

8(592 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

6 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

6 

76 

8808 

8814 

8820 

8825 

88:51 

a837 

8842 

8848 

8854 

8859 

6 

77 

8865 

8871 

8876 

8882 

8887 

8893 

S89<t 

8C»04 

8910 

8915 

6 

78 

8921 

8927 

8932 

8938 

8943 

8;i49 

Wt54 

8!H50 

89(55 

8971 

6 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

902.5 

5 

80 

9031 

90.r) 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

5 

81 

90S5 

<.»()<H) 

9096 

9101 

910(5 

9112 

9117 

9122 

9128 

91:33 

5 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

5 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

5 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

5 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

5 

86 

9:>15 

9350 

9355 

93(50 

9:5(55 

9370 

9375 

9:580 

9385 

9390 

5 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

5 

88 

9445 

9450 

9455 

94(50 

94(55 

<y69 

9474 

9479 

ft484 

9489 

5 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

95:38 

5 

90 

9542 

9547 

9552 

9557 

9562 

95(56 

9571 

9576 

9581 

9586 

5 

91 

9590 

9595 

9600 

9605 

'Mm 

9(514 

9619 

tt()24 

!)(528 

9633 

5 

92 

9(5;38 

9643 

9(547 

9652 

<Hi57 

9(5(51 

<K5(>6 

!H571 

9675 

9680 

5 

93 

fH585 

9689 

96SV4 

9699 

!)703 

9708 

9713 

9717 

9722 

9727 

5 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

976:3 

9768 

9773 

5 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

5 

96 

9823 

9827 

9832 

98:5(5 

9841 

9845 

9850 

9854 

9859 

9863 

5 

97 

98(38 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9<)03 

9908 

4 

98 

9i)12 

9917 

9921 

9926 

9930 

9J)34 

99:39 

9<»43 

9948 

9952 

4 

99 

<iO.-(5 

9'.i(;i 

9'.Ki5 

9960 

9<>74 

997S 

99S:! 

99«7 

9991 

99fH! 

4 

350 


LOGARITHMS 


[Ch.  XIV 


Ex.  2.    Simplify 


'-^43  +  5^278 


The  operations  of  addition  and  subtraction  cannot  be  performed  by 
logarithms.  We  may  find  5  v278  by  logarithms,  add  this  to  43,  and 
tlien  perform  the  remaining  operations  by  logarithms.  We  handle 
the  problem  as  if  all  the  numbers  were  positive  and  then  put  the 
proper  sign  before  the  answer,  which  in  this  case  is  negative. 


Ex.  3.    Find  V.0247 

Log  iCom  =  i  log  .0247  =::  1  (-  2  +  .3927) 
=  (-  .6667 +  .1309)  =-  .5358 

We  cannot  find  this  in  the  table,  since  it  is  negative.     Hence,  we 
write  in  another  form  : 

log  VaMI  =  1  (-  2  +  .3927)  =  i (8.3927  -  10) 

=  (2.7976  -  3.3333)==  2.4643  -  3  =1.4643 

We  may  do  this  in  another  and  simpler  way.  By  subtracting  1 
from  the  —  2,  and  adding  1  to  the  mantissa,  we  obtain  |(—  3  +  1.3927) 
=  I  +  .4642     Hence, 


log  \/.0247  =  1.4642,  whence  V.0247  =  .2912 


Ex.4.    Solve  3^  =  17. 

We  might  put  L  =  3^  and  try  various  values  of  x  and  plot  a  curve. 


X 

1 

2 

3 

5 

¥ 

¥ 

tt 

L 

3 

9 

27 

15..59 

20.51 

Thus,  we  would  find  an  approximate  value  without  logarithms,  but 

the  computation  is  very  long.     It  can  be  done  much  more  simply  by 

logarithms. 
^  3«  =  17 

log  3*  =  log  17 

z  log  3  =  log  17,  by  III,  §  165 

lo"- 17      1.2304 


log  3 


.4771 


2.579 


168]  LOGARITHMS  351 

EXERCISES   IV:    CHAPTER   XIV 
Compute  by  the  use  of  logarithms : 

1.  43  X  75.  20.    (99.43)^ 

2.  437  X  9.63.  21.   (1.01)=". 

3.  439  X  .0372,   — ■  22.    (1.04)'*". 

4.  243.7  X  .179.2.  33    ^609. 

5.  .8752  X  .01529. 

6.  1.002x3.075./ 

7.  20-685. 

8.  .257 -=-1.73. 


24.    V.  005726. 


25.  V125.3. 

26.  A/r072.  V 

27.  ^200. 


9.  .0024 -=- 1.034. 

10.  10.07-- 4.617.  28.  V.005. 

11.  25,680 -=- 152,980.  29.  ^f. 

12.  3  --  .002463.  30.  V\, 

13.  47  X  6.3  X  250.  ^^     483  x  .035  -  2.461  ^ 

14.  246  X  .0072  X  102.  "        5x4.6x3547 

15.  76  X  50.04  X  .06004.  33     642  x  3729  x  .0007 

1705  X  2.004  X  .4  ' 

546  X  .0001  X  .4040 


16.  1205  X  6|  X  54f 

17.  241x171x2464.  33. 


^'  2.02  X  .003  X  .2623 

18.    (1.47)'^ 

34    J46X3.5 


19.    (3.057)^  "•*•   \9l^a02 


•  v: 


362  LOGARITHMS  [Ch.  XIV 

35      '/1^47x2391  (4.007)t  x  (.003^) 

A/  14.62x24  *^-  ^-^g 


36. 


5  X  24  X  3.1  44.    (4)3^2_^3. 

V2  X  18  X  640 ' 


37.    v^x  3^x240. 


38.     Vl9-4X^xl-^2 


45.    \.38V95. 


46. 


|5A/a'\t^475. 


TT5 


976     3.95  47.    \16-v92. 


39. 


40. 


'^47j<^3,246_.  Vl^^yg^ 

\89.32x  34.95  *8.  _• 

V16-V92 

5020  V.  00437 

4^97:3       ■  49     ^247  X  V44 


41. 


392  -  V44 

wVn  X  43   ^  

247^5462  '  50     -31  + V(3]r- 4.3117. 

2.31 

„     (14)t  X  (129)'   y  ^ 

•    -^^I^iT-  51.    5^-V52^  4.6.19. 

Solve  examples  52  to  59  for  x. 

52.   3^  =  12.  55.    14"' =27.  ^o    otrj     5.4^-5 

08.     /4^  = 


53.  9—7.  56.    16- -3  =  7.  ^-^ 

X 

54.  17^  =  1.  57.   2^^=47.  59.    256=4.7^-1. 

60.  Find  the  compound  amount  of  $  100  for  15  years  at  4  %. 

61.  Find  the  compound  amount  of  $  1  for  100  years  at  61^. 

62.  What  amount  put  at  compound   interest   at   4  %  at  a 
child's  birth  will  amount  to  $  1000  when  he  is  21  years  of  age  ? 

63.  At  what  rate  at  compound  interest  will  $  1  amount  to 
$10  in  40  years? 

64.  In  how  many  years  will  $  1  amount  to  $  2.00  at  4  % 
compound  interest  ? 


108]  SUMMARY  353 


SUMMARY  OF   CHAPTER  XIV:   LOGARITHMS,  pp.  334-352 

Review  of  Exponents  :  restatements.     Exercises  I. 

§  162,  pp.  334-335. 

Temporary  Table  :  logarithms  as  exponents  of  a  chosen  base ;  fig- 
ure for  n  —  10^;  short  table  from  figure  and  computation, — 
base  10.  §  163,  pp.  335-340. 

Simple  Computation  by  Exponents  :  illustrations  of  use  of  exponent 
laws.     Exercises  II.  §  164,  pp.  340-341. 

Formal  Definitions:  base,  —  the  number  whose  exponents  are 
used;  logarithm,  —  the  exponents  that  produce  given  num- 
bers;  equivalence  of  n  =  10-^  and  log  n  =  L. 

Formal  Rules :  logarithm  of  product,  —  sum  of  logarithms;  loga- 
rithm of  quotient,  —  diiference  of  logarithms;  logarithm  of 
power,  —  power  times  logarithm.  §  165,  pp.  341-;342. 

Characteristic  :  integral  part;  increase  of  1  for  every  digit  place; 
judgment  of  characteristic  without  table. 

Mantissa :  fractional  part ;  independent  of  decimal  point ;  tables 
necessary.  §  166,  pp.  343-344. 

Four  Place  Table:  logarithm  of  given  number;  number  for  given 
logarithm  ;  interpolation.     Exercises  III.  §  167,  344-347. 

•  Computations :  illustrative  examples ;  tables  given  (pp.  348-349). 
Exercises  IV.  §  168,  pp.  347-352. 


APPENDIX 
NOTE  I.  DETACHED  COEFFICIENTS 

1.  Detached  Coefficients.  We  may  considerably  shorten 
the  labor  of  many  operations  in  polynomials  by  writing 
only  the  coefficients,  taken  in  their  natural  order  after 
the  polynomials  are  arranged  in  ascending  or  descending 
powers  of  some  one  letter. 

2.  Multiplication. 

Thus,  to  multiply  3  a^^  —  7  a;-  -f  5  .^'  —  6  by  2  a--^  —  4  a;  +  3,  we 
merely  write 

3-    7  +    5-6 

2-4+3 


6-14  +  10-12 
-12  +  28-20  +  24 

9-21  +  15-18 
6  -  26  +  47  -  53  +  39  -  18 
and  write  the  product 

6  x'-  -  26  X*  +  47  x3  -  53  x^  +  39  x  -  18, 

the  power  of  x  which  belongs  in  eacli  term  being  clearly  indicated  by 
tlie  position  of  the  term. 

3.    Division. 

Likewise,  if  we  are  to  divide 

Gx^-26a;^  +  47a;''-r)3ar'  +  39a;-18  by  2a--4.T  +  3, 

354 


DETACHED   COEFFICIENTS  355 


2-4  +  3  Divisor 


3-7  +  5-6     Quotient 


we  write  only  the  coefficients,  as  follows : 

Dividend :  6  -  26  +  47  -  53  +  39  -  18 
6  -  12  +    9 

-  U  +  38  -  53 

-  14  +  28  -  21 

10  -  32  +  39 
10  -  20  +  15 
-12  +  24-18 
-12+24-18 

0     Remainder 
Whence,  the  quotient  is   3  x^  -  7  x-  +  5  a;  -  6.      Compare  with    the 
multiplication  performed  above. 

The  student  must  be  extremely  careful  not  to  omit  terms 
even  when  the  coefficient  is  zero,  for  the  position  of  the  term 
indicates  the  degree. 

Thus,  x3  +  2  X  -  5  should  be  written  1  +  0  +  2-5,  not  1  +  2-5, 
for  1+2-5  would  represent  a;^  +  2  x  —  5. 

4.  Division  by  x  —  a.  Division  by  a  simple  binomial 
of  the  form  {x  —  a)  is  especially  easy. 

Let  us  divide   a^  +  2  .-c  —  5  by  x  —  3.     We  write 


Dividend :   1+0  +  2-5 
1-3 
+  3  +  2 
+  3-9 


1—3  Divisor 


1  +  3+11     Quotient 


11-5 
11  -  33 

28     Remainder 

1V1  z^  +  2  X  —  5        o  ,   o  ^  I    1 1    1      28 

A\  hence,  — ^^ =  x^  +  ox  +  li  H -• 

X  -  3  a;-  3 

Note  that  this  becomes  even  more  simple  if  we  write  none  of  our 
numbers  twice.     Thus,  we  may  write 

Dividend ;   1  +  0  +    2  -    5  |1  -  3     Divisor 
_  3  -    9  _  ,33 


Quotient:     1  +  3  +  11|+  28     Remainder 
which  gives  all  the  information  in  shorter  time.     This  shorter  form 

may  lie  used  only  when  the  divisor  is  of  the  form  x  ±  a. 


356  APPENDIX 

EXERCISES   I:    NOTE  I  —  DETACHED   COEFFICIENTS 

Perform  the  indicated  operations : 

1.  (oi^  +  2x'-5x  +  3)x(x  +  2). 

2.  (5x''-3x'  +  7)x(3x'  +  x-5). 

3.  (x  +  2x'  +  3x''  +  x')x{x-2x'  +  5x^). 

4.  (x-\-2  +  2x-^-3x-^)x{x  +  3-x-^). 

5.  (4:a^  +  2x-l)x(3ar  +  2x). 

6.  (3x'  +  2x^-5x  +  4)x{3x^-7). 

7.  (2x^-3x  +  7)x(x'-2x  +  1). 

8.  {x'  +  6x'  +  12x  +  8)^(x  +  2). 

9.  (6x2  +  8x  +  16)--(x-  +  4). 

10.  (8a^-36a;2  +  54x-27)^(2.x-3). 

11.  (12a^-14.T  +  16)--(6a;  +  5). 

12.  (14  a^  - 13  x'  +  27  x^ -  6)--(2  .x^  +  3  .t  -  7). 

13.  (25  ic-"  +  5  x-^  -  2  x-2  +  a;-^  +  5) --  (5  x'^  +  3  cc"'  +  2). 

14.  (3cc3  +  2x2_5a;  +  7)-f-(3a;-l). 

15.  (ar^4-a;--4a;  +  6)--(x  +  3). 

16.  {x*  -  Ax- -  5x  +  10)^(x-2). 

17.  [(.-k"  +  x^ -x-l)^{x-  1)] -- (x  +  1). 

18.  (r''  -\-7x"+lox  +  25)  ^  (x  +  5). 

19.  (x'  -  ar'  -  6  .TT  +  X  -  3)  H-  (a;  -  3). 

20.  [(^2  +  2  iC  +  1)  X  (.T  +  3)]  ^  (.T  +  1). 

21.  [(2  .r'  +  5  X-  + 12  x  4-  5)  - (2  .-K  +  1)]  - (3  x  +  5). 

22.  [(2  .r"  +  4  a;2  +  3)  (8  a;2  + 16  x  +  16)]  --  [(4  a;  +  4)  (x  +  3)]. 

23.  [(3  .r'  +  4  a;  +  7)  (3  x^  -  5)]  h-  [(a;  +  5)  (9  x'  +  2  .t  +  3)]. 

24.  [x-*  -I-  4  x-"  +  3  x-3  -  2  X-'  +  a;-i  +  5  +  3  a-]  X  (.t-'  +  3). 


NOTE  II.    REMAINDER  THEOREM;   FACTORING 

5.  Factor  Theorem.     If  a  polynomial 

P  =  Ax''  +  Bx''-^  +  ■■•  +Lx  +  N 

has  a  factor  of  the  form  x  —  a,  we  have  P  =  (x  —  a)  •  Q. 
Hence,  the  equation  P  =  0  is  equivalent  to  the  equation 
(x  —  a)Q  =  0 ;  since  x=  a  is  a  solution  of  this  equation 
(as  is  seen  by  actually  substituting  a  for  x),  we  may  say : 

I.  If  P  has  a  factor  (a;  —  a),  then  a  is  a  root  of  the  equa- 
tion P  =  0. 

Conversely^  suppose  a  is  a  root  of  the  equation  P  =  0. 
Dividing  P  by  a:  —  a,  we  should  get  some  quotient  Q  and 
some  remainder  72,  ivhere  R  is  a  number  independent  of  x : 

p=Q(^x-a}-\-  R. 

Now  set  x  —  a.  Since  a:  =  a  is  a  solution  of  the  equation 
P  =  0,  we  have  first  P  =  0  when  x  =  a.  Next,  {x  —  a)  is 
zero  when  x=a.     Hence,  the  above  equation  reduces  to 

0  =  0  +  P, 

that  is,  P  =  0.  In  other  words,  the  remainder  obtained 
by  dividing  P  by  a:  —  a  is  zero,  or  P  is  exactly  divisible 
by  x—a. 

II.  If  x  =  a  is  a  root  of  P  =  0,  then  P  is  divisible  by 
x  —  a. 

6.  Factors  of  x"±y".  These  facts  enable  us  to  factor 
in  many  cases.  For  example,  a;"  —  1  =  0  is  satisfied  by 
setting  x  =  l.  Likewise,  x^  +  1,  a;^  +  1,  etc.,  or,  in  general, 
x"  +  1  where  n  is  odd.,  is  divisible  by  a;  +  1,  for  the  equation 
a;"  +  1  =  0  (n  odd)  is  satisfied  by  setting  a;  =  —  1. 

Finally,  af'  —  l  is  divisible  by  a;  + 1,  if  n  is  even,  for 
x"  —  1  =  0  is  satisfied  by  a:  =  —  1. 

367 


358  APPENDIX 

In  like  manner, 

(1 }  x"  —  y"  is  always  divisible  by  x  —  y. 

(2)  x"  —  ^"  w  divisible  by  x  -\-  y  it  n  is  even. 

(3)  x"  +  y"  is  divisible  by  x  +  y  if  n  is  odd. 

Many  forms  may  be  factored  upon  this  basis.     Thus,  x^  —  y^ 
is  divisible  by  x  —  y.     By  actual  division,  we  find 

x'i-  ys  =  (x  -  y){x^-  +  xy  +  y'^). 

Check:  1  +  1  +  1  i.e.  x2  +  xy  +  y^ 

1  —  1  i.e.  ^  —  y 

1  +  1  +  1 
—  1  —  1  —  1  multiplied 

1  +  0  +  0-1  xg  -y^ 

We  find  also,a:8  j^  yZ  -  (^^  +  y)(x'^  -  ^y  +  y^)  ; 

X^-y^=   [(2-2)2  _  (y'i^oj   ^   (^.-2  _   ^.2)  (^.-2  +  ^2) 

=  (x-y)(x  +  ;v)(x2  +  y2). 

3,5  _  j,5  _   (2;  -  y)  (X*  +  X3y   +  X-y2  _f.    ^.yS  _!_  ^4)  . 

a:^  +  y^  =  (x  +  //)(a-^  -  :r^//  +  x'^y'^  -  xy^  +  ?/*); 

X6  -  ^6  =   [(X3)2  -  (^3)2]  =   (x3  -  y3)(3.3  +  ^^S) 

=  (x  -  y)(x2  +  xy  +  //2)(x  +  y)(x2  -  x.y  +  y2); 

Or,alsO,  X6  -  ?/«  -   [(x2)3  -  -   (y2)3]  =  (^.-2  _.  ^2')  (^4  +  ,2,/2  +  ^4) 

=  (x-y)(x  +  //)(x''  +  xV+2/'); 

whence,  comparing  with  the  preceding,  we  find : 
X*  +  x2y2  +  ?/4  ^  (x2  +  xy  +  y2)(x2  _  x//  +  y^), 

x'  ±y''  ={x±y)  (x6  :p  x^y  +  x^"  ^  x-^y^  +  x^y"  ^  xyS  +  y«) 
and  so  on. 

These  forms  may  be  used  as  type  forms,  and  other  ex- 
pressions may  be  factored  by  comparison  with  them,  as  in 
Chapter  IV,  p.  91.  See  also  Chapter  XIII,  p.  331,  where 
the  same  results  are  found  by  a  different  method. 


REMAINDKR   THEORExM.     FACTORING  359 

EXERCISES   I:    NOTE    II  — FACTOR   THEOREM 

Factor  the  following  expressions: 

1.  x-*-l.  4.    a-'^-64.  7.    of* -27.        10.    ar*— a'. 

2.  x'+l.  5.    x-^-lC.  8.    .r'-32.        11.    x'  +  a^ 

3.  8x^-1.  6.    C4.i--16.       9.    .i-«-2/».  12.    a-"-r". 

13.  8(a;  +  ?/)='+(.f-^)'.  17.    27 /-^Z  -  8. 

14.  (.T  +  2//--(.f-?/)^  18.    (.^•  +  2/)'-(-^--2/)'- 

15.  (x-yf-{x  +  yy.  19.    81iJ^-62o5^ 

16.  (.r  + 1)^  +  1.  20.    m^/i'"  — mV. 

7.  Factors  of  Polynomials.  Some  polynomials  may  be 
factored  by  the  above  principles. 

Ex.  1.  To  factor  P  =  x*  -  4  x^  -  x"^  +  16  x  -  12,  we  may  try  the 
factors  of  the  last  term  -  12,  since  -  12  must  be  the  product  of  the 
constant  terms  in  all  the  factors.  Try  ±  1,  ±  2,  ±  3,  ±  4,  ±  6,  ±  12. 
Thus,  to  try  +  1  we  set  x  =  +  1  in  P;  this  gives  P  =  (for  x  =  1) 
1  -  4  -  1  +  16  -  12  =  0;  hence,  x  -  1  is  a  factor  of  P.  If  we  try 
all  the  other  numbers  above,  we  shall  find  that  the  factors  of  P 
are  (x  -  1),  (x  -  2),  (x  +  2),  (x  +  3).  Check  this  by  multiplying 
these  factors  together. 

EXERCISES    II :    NOTE   II  —  FACTORS    OF   POLYNOMIALS 

Find  by  the  factor  theorem  a  factor  of  the  left  side  of  each 
of  the  following  equations  and  by  so  doing  find  one  real  root 
of  each  equation : 

1.  4:a^-5x  +  l  =  0.  6.  x*-10ar-20x-16  =  0. 

2.  af'-ar~.r-15  =  0.  7.  .r-« -4ar- 16.T  + 15  =  0. 

3.  4:x'  +  16x'-9x-6^=0.  8.  x^-Sx-  +  12x-5^0. 

4.  ar'-2a;2-.^•  +  2  =  0.  9.  x^ -x -Sx- +  3  =  0. 

5.  X* + 3  X-  -  6  .r  +  3  =  0.  10.  x'  +  x  -  5  x'' -  5  =  0. 


360  APPENDIX 

8.  Remainder  Theorem.  The  work  in  §  5  proves  an- 
other interesting  fact.     For  we  had 

where  Q  stands  for  the  quotient  and  R  for  the  remainder 
upon  dividing  P  hy  x—a.  Suppose  P^j-gj.^^^^^0.  Then 
substituting  x~a,  we  get  P(jg^j.^a)'=  Ri  since  the  term 
Q(x  —  a)  certainly  falls  out  when  x  =  a.     In  other  words  : 

III.  The  remainder  found  upon  dividing  P  hi/  x  —  a  is 
equal  to  the  value  of  P  when  x  is  replaced  hy  a. 

In  long  expressions  it  is  easier  to  divide  and  find  R 
than  to  substitute  a  in  the  place  of  x  in  P. 

Thus,  to  find  the  value  of  P=a/'^— 5a;''+7a;+3,  when  a;  =  4, 
we  may  write       p^^^^  =43-5  (4^)  +  7-4  +  3 

and  actually  compute  4^,  5  (4-),  etc. ;  or  we  may  divide  P  by  a;— 4 


and  find  R : 


a;3_  5x2  + 7x+    .six- 4 

x^  —  4  x^  I  a;2  —  X  +  3 

—  x^  -\-  1  X 

—  a;2  +  4  X 


3a;+    3 
3x-  12 

15  =  i2 

Then,  P(^=4)  =  (4)3  -  5(42)  +  7  .  4  +  3  =  15^  which  the  student  may 
verify  by  actually  computing  P(x=iy 

The  work  is  even  shorter  in  detached  coefficients,  see  p.  355. 

EXERCISES    III  :    NOTE   II  —  REMAINDER    THEOREM 

Find  by  the  remainder  theorem  the  value  as  shown  in  each 
of  the  following  examples.  Verify  the  first  three  by  actually 
substituting  the  indicated  value  for  the  unknown. 

1.  p=:  x^  _  5  a;2  + 15  X  —  75,  for  x  =  5. 

2.  P=a^-27a;*  +  15x'2-35a;  +  125,  for  a;  =  -3. 

3.  P  =  x'-^ox  +  n,iov  x  =  2. 

4.  P=x^- 15  X*  +  25  a^ - 125 a!^  +  50,  for  a;  =  -  4. 

5.  P-^x^-3x^  +  15x-  20,  for  a;  =  3. 

6.  P=  x^  +  4 ar' -  12 a;=^  + 1  =  0,  for  x  =  -2. 


NOTE  III.    CHOICE   AND   CHANCE;   PERMUTA- 
TIONS  AND    COMBINATIONS 

9.  Choice.  The  arrangements  of  objects  and  the  possi- 
bilities of  choice  form  the  basis  of  this  note. 

As  a  typical  example,  suppose  that  I  wish  to  select  a  route  from 
Chicago  to  Liverpool,  via  New  York,  from  amoug  four  railroads  from 
Chicago  to  New  York  which  I  would  consider,  and  three  lines  of 
steamers  from  New  York  to  Liverpool.  Having  taken  any  railroad 
to  New  York,  I  may  go  to  Liverpool  on  three  different  lines.  Since  I 
have  four  railroads  from  which  to  choose,  the  total  number  of  ar- 
rangements is  4  X  3,  or  12. 

In  general,  if  one  choice  is  made  in  n  ways  and  then 
another  independent  choice  made  in  m  ways^  the  total  num- 
ber of  arrangements  for  the  two  choices  is  n  x  m,  or  nm. 

Likewise,  for  any  series  of  choices,  the  total  arrange- 
ments of  all  choices  is  the  product  of  the  separate  numbers 
for  the  separate  choices. 

10.  Chance.  The  chance  of  selecting  a  given  object 
from  among  a  number  of  objects  decreases  as  the  number 
of  objects  increases.  If  among  live  balls  in  a  bag  there 
is  one  white  one,  the  chance  of  selecting  the  white  one  at 
a  random  choice  is  1  to  5,  or  \.     In  general,  the  chance  of 

selecting  one  special  object  among  n  objects  is  -  • 

The  chance  increases  if  there  are  more  favorable  possibilities.  If 
there  are  tivo  white  balls  among  five,  the  chance  of  drawing  a  white 
ball  is  twice  as  great  as  it  is  if  there  is  only  one,  i.e.  the  chance  is  |. 

In  general,  the  chance  of  selecting  one  of  m  objects  out 


m 


of  a  total  of  n  objects  is 


361 


362  APPENDIX 

EXERCISES  I:    NOTE  III  — CHOICE  AND  CHANCE 

1.  Two  doors  in  one  room  and  three  doors  in  another  (not 
adjoining)  room  open  on  td  a  common  court.  In  how  many 
ways  may  one  go  from  one  room  to  the  other  ? 

2.  A  man  has  three  coats,  two  vests,  and  five  pairs  of 
trousers.     In  how  many  ways  can  he  dress  ? 

3.  A  boy  can  go  to  school  by  five  different  roads.  In  how 
many  ways  may  he  go  and  return  ?  In  how  many  ways  can 
he  arrange  his  trips  on  six  different  days  ? 

4.  There  are  eleven  horses  in  a  pasture  and  nine  saddles 
in  a  barn.  How  many  choices  of  saddle  and  horse  may  be 
made? 

5.  What  is  the  chance  of  selecting  a  black  ball  out  of  a  bag 
containing  five  black  balls  and  one  white  ball  ? 

6.  A  bag  contains  six  grains  of  white  corn  and  five  grains 
of  yellow  corn.  What  is  the  chance  of  selecting  a  white  grain 
the  first  time  a  grain  is  taken  ?  If  a  white  grain  is  drawn  the 
first  time  and  kept  out,  what  is  the  chance  of  getting  another 
white  grain  on  the  second  drawing  ?  What  is  the  total  chance 
that  a  white  grain  will  be  drawn  both  times  ? 

7.  What  is  the  chance  of  throwing  "  heads  "  in  tossing  a 
coin  ? 

8.  If  three  coins  are  thrown  up  together,  what  is  the  chance 
that  at  least  one  will  fall  "  heads  "  ?  What  is  the  chance  that 
two  will  fall  "heads"? 

9.  Dice  are  usually  cubes  marked  on  the  six  faces  with 
the  numbers  from  1  to  6.  What  is  the  chance  of  throwing  a 
"2"  with  one  die?  What  is  the  chance  of  throwing  a  "2" 
with  two  dice?  What  is  the  chance  of  throwing  a  double 
"  2 "  {i.e.  a  "  2  "  on  each  die)  with  two  dice  ?  What  is  the 
chance  of  throwing  two  numbers  whose  sum  is  ten  with  two 
dice  ? 


CHOICE   AND   CIIAXCE  363 

11.  Permutations.  The  number  of  arrangements  of  a 
given  set  of  objects  in  order  is  called  the  number  of  permu- 
tations of  them. 

Thus,  given  five  pictures  to  be  hung  upon  a  wall,  any  one  may  be 
placed  at  the  extreme  right,  then  any  of  the  four  remaining  ones 
next,  then  any  one  of  the  three  remaining  ones  next,  then  any  one 
of  the  two  remaining  ones  next,  and  finally  the  last  one  must  be 
liung  at  the  extreme  left.  The  number  of  possibilities  is  given  by 
§  9.  There  are  five  distinct  choices;  the  first  with  5  objects  from 
which  to  choose,  the  second  with  4,  and  so  on  ;  hence,  the  total  number 
of  possible  arrangements  is  5  x  -i  x  3  x  2  x  1  =  120. 

In  general,  if  n  objects  are  to  be  arranged  in  an  order, 
any  one  of  the  n  objects  may  be  put  first,  then  any  of  the 
i-emaining  n—1  next,  then  any  of  the  remaining  n  —  2 
next,  and  so  on  to  the  last.  The  total  number  of  possible 
arrangements  is  n  {n  —  1)(h  —  -)  — t  •  3  •  2  •  1. 

12.  Permutations  among  a  Limited  Number.  If  less 
than  the  whole  number  of  objects  are  to  be  arranged  in 
order,  the  number  of  possible  arrangements  is  evidently 
reduced. 

Thus,  if  we  desire  to  select  four  out  of  ten  candidates  for  an  office 
and  arrange  them  in  order  of  merit,  the  number  of  possible  arrange- 
ments is  smaller  than  if  all  ten  were  to  be  arranged  in  order  of 
merit.  Any  of  the  ten  may  take  first  rank,  any  of  the  nine  remain- 
ing second  rank,  any  of  the  eight  remaining  third  rank,  any  of  the 
seven  remaining  fourth  rank;  but  here  we  must  stop.  The  total 
number  of  arrangements  (or  permutations)  is 

10x9x8x7  =  5010 ; 

whereas,  if  all  ten  were  to  be  arranged  in  order,  the  total  number  of 
arrangements  (or  permutations)  would  be 

10x9x8x7x6x5x4x3x2x1=  3,628,800. 

In  general,  if  from  among  n  objects  we  are  to  make  an 
arrangement  of  7n  objects  in  an  order,  we  may  choose  any 
one  first,  then  any  of  the  remaining  n  —  1  second,  and  so 
on  for  m  steps.     The  last  choice  will  be  w  —  w  +  1  objects. 


364  APPENDIX 

since  n  —  m  will  still  remain  after  the  last  choice.     The 
number  of  arrangements  is,  therefore, 

n(n  —  l)(w  —  2)  •••  (n  —  m-\-  2)(n  —  m+  1), 

13.  Factorial  Notation.  A  convenient  notation  for  the 
kind  of  expressions  just  found  consists  in  writing  2!  for 
2  •  1  ;  3 !  for  3.2.1;  4  !  for  4.3-2.1,  etc. ;  in  general, 

nl  =  n(n-l)(n-2}---4:-S-2-l. 

The  sign  n  !  is  read  "  factorial  /?." 

The  permutations  of  n  objects,  m  at  a  time,  is  often  de- 
noted by  P„^  ^.  If  m  =  w,  all  the  objects  are  to  be  arranged  ; 
we  then  write  P„,  „  for  the  number  of  permutations. 

Using  this  notation,  the  results  found  above  are 

-Pn,n  =  nl  and  P„, ,„  =  - —  • 

(w  —  m) ! 

EXERCISES  II:  NOTE  III  — FACTORIALS  ;   PERMUTATIONS 

Compute  the  value  of : 
1.  5!      2.  6!      3.  12!     4.  (8!)-(4!).      5.  (5 !  x  4!)-(G !). 

6-      Ps,  1  ;     -P5,  2  ;     P7,  4  5     -^  8,  H  ;     -'15,  5  5     -'  M,  4  5     -*   n+l,  h-1  i      -Pn+k,  k  5      -*   12,  V)- 

Write  in  abbreviated  form  and  compute  the  number  of  per- 
mutations of : 

7.  8  objects  taken  3  at  a  time. 

8.  15  objects  taken  7  at  atime. 

9.  h  objects  taken  1  at  a  time. 

10.  n  +  1  objects  taken  n  —  1  at  a  time. 

14.  Combinations.  If  we  merely  wish  to  select  objects 
from  among  a  given  set  of  objects  without  arranging  them 
in  order,  many  of  the  permutations  become  equivalent. 

Thus,  if  among  the  ten  candidates  of  the  problem  in  §  12  we  wish 
to  select  four  without  placing  them  in  order  of  merit,  the  same  four  men 
would  be  arranged  in  4  !  =  4  .  3  •  2  . 1  =  24  different  arrangements  in 
§  12 ;  these  arrangements  are  all  equivalent  if  we  do  not  specify  the 


PERMUTATIONS    AND   COMBINATIONS  365 

order.     Hence,  the  number  of  combinations  {i.e.  not  counting  differ- 
ent arrangements  of  the  same  ones)  is  2V  of  the  previous  number,  i.e. 
10- 9- 8 -7 


^  4.3-2-1 


=  210. 


In  general,  if  C„^^  represents  the  number  of  combina- 
tions (i.e.  selections  not  counting  different  arrangements 
of  the  same  objects)  among  n  objects  chosen  m  at  a  time, 


f,       _  Pn.  m  _  (n  -  m) ! 


^m,m  m\  (n-m)lml 

EXERCISES  III:   NOTE  III  —  COMBINATIONS 

1.  Find  the  value  of  (7  !)--(4  !  x  3  !). 

2.  Find  the  value  of  C7, 3 ;   C5, 4 ;   C5, 5. 

3.  Find  the  values  of  (74,2;  A,  25  hiTd  C4, 2 -J- -P4, 2- 

4.  Find  Ps^sj  Ps,5]  hence,  find  Cg^^. 

5.  If  a  farmer  has  twelve  horses,  how  many  different  teams 
of  two  horses  each  may  he  use  ? 

6.  How  many  different  and  distinct  committees  of  five  may 
be  selected  from  a  group  of  15  men  ? 

7.  In  how  many  ways  may  3  books  be  selected  from  12 
books  ? 

8.  In  how  many  ways  may  the  sum  seven  be  thrown  with 
two  dice  marked  on  the  faces  with  the  numbers  from  1  to  6  ? 
With  three  dice  ? 

9.  In  how  many  ways  may  three  debaters  be  chosen  from 
a  squad  of  18  ?     How  many  if  six  must  be  chosen  ? 

10.  In  a  plane  are  ten  points,  no  three  of  which  are  in  a 
straight  line.  How  many  triangles  may  be  formed  with  three 
of  the  points  as  vertices  ? 

11.  In  how  many  ways  can  four  men  and  three  women  be 
selected  from  eight  men  and  seven  women? 


NOTE    IV.     INEQUALITIES 

15.  Operations  on  Inequalities.  We  have  used  a  few 
inequalities,  but  we  have  not  worked  with  them  systemati- 
cally. The  signs  <  (read  "less  than")  and  >  (read 
"  greater  than  ")  are  already  known.  A  few  statements 
that  will  be  understood  at  once  are  now  given : 

(1)  11  a>b^  then  b<ca. 

(2)  If  a >  ^  and  b>c,  then  a^o. 

(3)  \i  a^h  and  5  >  c,  then  a'>c.  (^  is  read  "  greater 
than  or  equal  to,"  and  <  is  read  "less  than  or  equal  to.") 

(4)  \i  a^h  and  b  <c^  then  a<c. 

(5)  If  «  >  5,  then  ka  >  kb  if  k  >  0. 

(6)  If  a>b,  then  ka<kb  ifk<0,  for  changing  the  sign 
evidently/  reverses  the  inequality. 

(7)  If  a  >  5  a7id  c  >  d,  then  a-\-  c>b  -\-  d. 

(8)  If  a>b,  then  a  ±  x>b  ±  x  where  x  is  any  number. 

These  rules,  together  with  a  clear  understanding  of 
what  is  intended,  enable  us  to  woi'k  with  all  ordinary 
inequalities. 

Ex.1.    For  what  values  of  cc  is  2  a;  — 3  >  6  — ic? 

Subtracting  6  —  a;  from  each  side,  we  get 

(2  a;  -  8)  -  (6  -  x)  >  0, 

or,  3a;-9>0. 

Add  9  to  each  side :  3  x  >  9. 

Divide  both  sides  by  .3  :  a:  >  3. 

Hence,  if  x  >  3,  then  2  a;  -  3  >  6  -  x. 

366 


INECjlALITIES 


36" 


1 

N 

rori 

\ 

/ 

\ 

^ 

/ 

- 

\ 

6' 

y 

S-^ 

4 

/ 

\ 

fj 

\ 

/ 

\ 

V 

/ 

( 

\ 

/ 

k 

1 

\ 

1 

s 

/ 

\ 

/ 

\ 

/ 

\ 

Fig.  72. 


16.  Graphical  Solutions.  The  prol)lGm  just  solved  may 
l)e  clone  grapliically.  Tims,  let  ?  =  22;  —  3, r  =  6  —  a:  where 
r  and  I  denote  the  right 
and  left  sides  of  tlie 
above  inequality.  Plot- 
ting each  of  these  on 
squared  paper,  we  have 
two  straight  lines,  as 
shown.  It  is  clear  from 
such  a  figure  that  ^  >  r 
whenever  the  line  I  =  2x 
—  3  is  above  the  line 
r=  G  —  X.  This  happens 
evidently  for  all  values  of 
X  greater  than  a;  =  3,  for 
the  lines  cross  at  x=  S,  and  they  surely  do  not  cross  again. 

We  may  use  this  fact  to  advantage  in  harder  examples.     Let  us 
find  the  values  of  x  for  which  x^  +  2  x  -  8  >  0.     Call  I  =  x^  +  2x  -  S 

the  left  side  and  draw  the  figure  ;  it 
is  as  shown  (compare  pages  183,  204). 
We  see  that  I  >  0  when  the  curve  is 
above  the  main  horizontal ;  this  hap- 
pens twice,  once  to  the  left  of  x  =  —  4, 
once  to  the  right  of  x  —  +  2.  These 
points  are  to  be  found  by  solving  the 
equation  1  =  0,  i.e. 

x2  +  2  X  -  0, 

of  which  the  solutions  are 

X  =  +  2,  X  =  -  4. 

Likewise, 

l<0,i.e.  x2+  2x-8<0 

for  all  values  of  x  between  x  =  —  4  and 

X  —  4-  '^.     ■ 
Fig.  73.  x  _  -t-  -. 


I 

\     l-o 

1 

-5     4 

r 

4  ^ 

r 

.     1 

t 

t 

l>o\ 

t 

i        0 

I       ? 

L_ 

J 

I 

I 

4  '^ 

t 

/i!=x^+2x-  8 

\ 

/ 

V  n 

7 

^  ^ 

r 

^.>^ 

368  APPENDIX 

Similarly,  in  any  case  we  may  represent  the  left  and 
right  sides  graphically  and  see  when  one  exceeds  the 
other.  It  is  evident  from  a  figure  that  we  should  first 
find  when  the  two  sides  are  equal,  for  the  points  for  which 
this  is  true  are  points  of  intersection  of  the  graphs. 

EXERCISES   I:    NOTE   IV  — INEQUALITIES 

Write  down  in  the  following  examples  the  conclusions  you 
could  draw  from  what  is  given,  as  directed: 

1.  10  >  6 ;  multiply  both  sides  by  5 ;  by  —  5 ;  by  ^ ;  by  —  |^ ; 
divide  by  2 ;  divide  by  \ ;  add  4  to  each  side ;  add  —  4 ;  add 
—  20 ;  subtract  4 ;  subtract  6.  (Perform  each  operation  on  the 
given  inequality  only.) 

2.  - 10  <  -  6 ;  multiply  by  +  5 ;  by  -  5 ;  by  i ;  by  -  ^ ; 
add  10  to  each  side ;    subtract  4. 

3.  3>2;  multiply  by  x  (if  x  is  positive);  divide  by  x 
(positive) ;  add  x  to  each  side;  subtract  x. 

4.  3  >  2 ;  multiply  by  x  (if  .t  <  0) ;  add  x ;  subtract  x. 

5.  Given  3a;  —  2>a:  +  4,  subtract  x  from  each  side ;  then 
add  2  to  each  side  of  the  resulting  inequality;  then  divide 
both  sides  by  2. 

Draw  the  figures  and  find  the  values  of  x  for  which : 

6.  3  a;  -  2  >  a;  4-  4.  9.   a;^  >  6  —  x.  12.    a;^  +  4  a;  -  5  >  0. 

7.  .T  -  2  >  3  —  a:.  10.  x">bx  —  4.        13.   .r^  —  2  x  —  3  >  0. 

8.  4x-2>2x  +  3.     11.  x'Kx''.  14.   ar'  +  2a;2-3>0. 


NOTE   V.     THE   BINOMIAL   THEOREM 

17.  Formula.  We  have  learned  how  to  write  down 
certain  powers  of  binomials  by  inspection.  (See  §§  57,  62, 
pp.  93,  99.)     Thus,  we  had 

(1)  (a  +  ^)2=a2+2a6  +  52(p.  93), 

(2)  (a  +  ly  =  a3  +  3  a%  +  3  a52  +  ^3  (p.  99^^ 

(3)  (a  +  hy  =  a^  +  \an  +  (j  (fi}P-  +  4  ai^  +  6*  (p.  99), 

and  so  on.  We  shall  now  prove  the  following  formula, 
called  the  binomial  theorem,  which  gives  the  result  for 
any  positive  integral  value  of  the  exponent  n : 

U*'  term)  (Sd  term)  (3d  *erm) 

(4)  (a  +  hr  =     a"     +   na'^-^h  +  ^^^  ~  ^^  al'-W 

^    I 

(A//fc  ierm) 
7l(w-l)(n-2)^„_3^3  (  student  tcrite^ 

3  f  \      5th  term      J 

{rth  term) 

,  n(n-V)(n-2')  •••  (n  -  r  +  2)  ^^-r+ijr-i 
(r-1)! 

{(r+  Dthierm) 

^  n(n-l)(n-2)  •••  (w  -  r  +  1)  ^„_,^,  ^    ^^ 

r  ! 

(nt/i  terrn)      (last  term) 

where  r  I  means  1  •  2  •  3  •••  (r  —  1)  •  r,  as  on  p.  364. 

To  prove  this  formula,  we  shall  show  that  if  it  holds 
for  any  positive  integral  value  of  n,  it  holds  also  for  the 
next  higher  value  of  n.      For  this  purpose,  assume  for  a 

369 


370  APPENDIX 

moment  that  (4)  is  correct  and  multiply  both  sides  by 
(a  +  6)  ;   this  gives 

(rt  X  lut  term)  {h  x  Isf  term)  (a  x  2d  term) 

(5)    (a  +  6)«+i=        [a"+i]         +  [a"5       +       na"b] 

(b  X  Sd  term)  (n  x  3d  term)  (h  x  3d  term)  +  {ax  Uth  term) 

+     na     0   -^    2l       a     f*      -t-   I    this  doivn    J  ^ 


+ 


(6  X  rth  term) 

'n(n-l')(7i-2)  ■■■  (n-r  +  2') ^„-r+i5r 

(fl  x(r  +  l)th  term) 

+  n(n-l)(n-  2)  •••  (/i-r+1)  ^„-r+i^,;1  ^ 


{bxnterm)  (a  x  Inxt  term)  (b  x  hint  term) 

+    [waJ"     +       a  J"  J        +     [^>"+i]  ; 

or,  collecting  the  terms,  we  have 

(a  +  J)n+i  =  a"+i  +  (w  +  l)a"i  +  ^^-^p^ a«-i52 ^.  ... 

^  (n  +  l)n(n-l)(in-2)  -(n-r+2)^^_,^-^^,  ^   ... 

r ! 
^-(n  +  l)ai"  +  ^"+^ 

which  is  the  same  as  (4)  with  (w  +  1)  put  in  place  of  n. 
It  follows  that  if  formula  (4)  holds  for  any  positive  inte- 
gral exponent  w,  it  holds  also  for  the  next  higher  iiitegral 
exponent  n -\- \  \  for  we  have  derived  (5)  on  the  assump- 
tion that  (4)  is  correct.  Now  we  know  (4)  holds  if  w  =  2, 
for  if  w  =  2,  the  formula  reduces  to  (1),  which  we  know 
is  correct.  Hence,  the  formula  must  also  hold  if  n  =  3, 
by  our  argument  just  given.  Since  it  holds  for  n  =  3,  it 
must  hold  for  w  =  4 ;  since  it  holds  for  w  =  4,  it  must 
hold  for  n  =  5  ;    etc.     In  fact,  we  may  say  that  the  formula 


BINOMIAL    THEOREM  371 

(4)  holds  for  any  positive  integer  n  whatever^  for  we  should 
eventuully  reach  any  given  one.  Thus  the  formula  is 
proved. 

The  style  of  argument  just  used  is  called  mathematical 
induction,  and  it  is  useful  in  many  other  proofs. 

A  proof  by  mathematical  induction  shows:  (1)  that  a  fonniila  is 
true  for  at  least  one  particular  integral  value  of  one  of  the  letters  ; 
then,  (2)  that  if  this  formula  is  true  for  a  given  integral  value  of  that 
letter,  it  is  true  also  for  the  next  higher  integral  value;  from  (1)  and 
(2)  it  follows  that  the  law  is  true  when  the  letter  mentioned  has  any 
integral  value  greater  than  the  value  actually  tested.  This  reasoning 
may  also  be  modified. 

18.  Notes  and  Examples.  In  the  formula  just  given  it 
should  be  noted  that : 

(1)  There  are  (n  +  1)  terms  in  all;  e.g.  there  are  6 
terms  in  (a  -f-  h^. 

(2)  The  exponent  of  a  is  reduced  hi/  one  from  each  term 
to  the  next  one  ;  the  exponent  of  h  is  increased  by  one. 

(3)  The  coefficients  are 


1; 

5,   1; 


[Let  the  student  extend  this  table.  A  rule  is  readily  formed  for 
obtaining  any  of  these  numbers,  —  by  adding  any  two  successive  ones 
in  the  same  row  we  get  the  number  directly  below  the  second  one  of 
the  two  added.] 

Ex.  1.    (a  +  6)-'=a^+4  a'-^h  ^"^^^^ a'-^h^ 

4(4-l)(4-2)   ,_3^3  ,  4(4-l)(4-2)(4-3)^,_,^, 
"^        1.2-3  1.2.3.4 

=  „4  _,_  4  ^-^^l^  _f.  6  „2^2  _,_  4  ^^^/  _,_  jji 


for  w  =  1 

1; 

for  7i  =  2 

9 

1; 

for  n  =  3 

3, 

3, 

1; 

for  n  =  4 

4, 

6, 

4, 

for  71=  5 

5, 

10, 

10, 

for  n=  6 

6, 

15, 

20, 

372  APPENDIX 

This  result  agrees  with  (3)  above.'  Also  that,  if  we  did  not  know 
where  to  stop,  we  should  find  the  next  term  exactly  zero  if  we  did 
write  it  down ;  hence,  there  is  no  danger  of  writing  too  many  terms. 

Ex.  2.    (a -\-hy  =  a'  +  l dJ-^h  +  ^-^^-^^ ci'-'b'  +  •  •  •    (student 

complete),  =a^-\-7  a%  +  21  a^h-  -\ (student  complete). 

Ex.  3.    {2x  +  ^yY={2xy  +  b{2x)\^y)  +  ^{2x)\3yf 

•  +fi|^(2^)X3y)^  +  ^;^;^:^(2a^)(3y)^  +  (3y)- 

=  ?,2x'  +  240  .tV  +  720  :i?f  + 1080  x-f  +  810  x^f  +  243  f. 

Ex.4.  (2aj-32/)«  =  [2.x  +  (-3?/)]''  =  (2a;)«  +  6(2x)^(-3^) 

+  ^(2x)X-32//  +  f4^(2x->X-32/)^+-  (student  com- 

plete),  =  64  .-c^-STG  x>y+1\.m  x'y--  4320  xh/  +  •••• 

Ex.  5.  The  6th  term  of  (a  +  6)>"  is  given  by  putting  n  =  12 
and  ?•  =  6  in  the  9-th  term  of  (4) : 

6th  term  of  (a  +  6)'^  =  ^\'^]'l^\^:^ a'b'  =  792 a'h^ 

EXERCISES:  NOTE   V  — BINOMIAL   THEOREM 

Write  out  in  full : 

1.  {a-\--b)\  4.   i^x^yy.  7.   (l+a^)«. 

2.  (3m  +  2w)*.  5.   (2.^--3?/)^  8.   {mp-Vf. 

3.  (a;-2/)^  6.   (4^2  +  ^^)*.  9.    {I's-ty. 

Write  out  the  first  three  terms  of : 

10.  {x-yj\  12.    (a +  6)^.  14.    (1  -  ^r^". 

11.  (4.^2 +  3  2/^)*.         13.   (kv  +  1)'.  15-   (2  +  4a)^ 

16.  Write  out  the  6th  term  of  (a  +  h^. 

17.  Write  out  the  5th  term  of  (1  -f  .t)". 

18.  Write  out  the  12th  term  of  (2  a  -  3  h^. 


NOTE   VI.     EUCLIDIAN   METHOD      H.C.F. 
AND    L.C.M. 

19.  Euclidian  Method.  In  Chapter  V,  p.  118,  we  defined 
H.C.F.  and  L.C.M.,  and  showed  how  to  find  them 
if  the  given  expressions  coukl  be  easily  factored.  The 
following  method  applies  to  all  polynomials,  no  matter 
how  difficult  the  factoring  may  be. 

Ex.  1.   Find  the  H.C.F.  of  ^ 

^  =  3x-2-llx  +  6  and  5  =  12 x3+a;2_i2x4-4. 
Divide  the  expression  of  higher  degree  (fi)   by  the  other  C^)" 


3x2-  llx  +  6  =  ^ 


4x  +  15  =  Quotient 


£  =  12  x3  +       x2  -    12  X  +    4 
12x3-44x2+    24  X 

45x2-    36x+    4 
45  x2  -  165  X  +  90 

129  X  -  86  or  43  (3  x  -  2)  =  Remainder 

Calling  the  quotient  Q  and  the  remainder  R,  we  have,  as  always, 

B=  Q-  A  +  R 

i.e.  12  x3  +  x2  -  12  X  +  4  =  (4  X  +  15)(3  x2  -  11  x  +  6)  +  43  (3  x  -  2). 
Any  factor  of  both  A  and  B  must  also  be  a  factor  of  the  remainder  R, 
for  R  =  B  —  QA,  so  that  a  factor  of  both  A  and  B  is  a  factor  of 
B  —  QA,  ivhich  is  nothing  but  R.  For  example,  a  factor  of  12  x°  +  x2 
—  12  X  +  4  and  3  x2  —  11  x  +  6  must  also  be  a  factor  of  43  (3  x  —  2). 

All  common  factors  of  B  and  A  are  also  common  factors  of  A 
and  R,  and  vice  versa.  Hence,  we  may  take  instead  of  the  given 
problem  the  simj)ler  one :  to  find  the  common  factors  of  A  and 
R;  i.e.  of  3  x^  - 11  x  +  6  and  43  (3  x  -  2). 

This  new  problem  may  be  done  by  inspection,  or  if  this  is  too  diffi- 
cult in  any  case,  by  repeating  the  process.  In  our  example  it  is  clear 
that  3  X  —  2  is  a  common  factor;  it  is  also  the  only  one,  by  the  argu- 
ment just  used.     Hence,  the  required  H.  C.  F.  is  3  x  —  2. 

373 


374  APPENDIX 

Ex.2.   Find  the  H.C.F.   of  A  =  x' +  4:a^-j-2x' +  x-2  and 

B  =  2a^-\-5x* -7x^-2  x'-5x-\-7. 


Dividing  as  above,  we  find 

B  =  2x^  +  5x*-    7x3-2a:2-5x  +  7 
2x5  +  8x^4-    4x^  +  2x^-ix 

-  3  X*  -  11  x8  -  4  x2  -     x  +  7 

-  3  x4  -  12  x3  -  6  x2  -  3  X  +  6 


x^  +  4  x3  +  2  x2  +  X  -  2  =  ^ 


2  X  —  3  =  Quotient 


x^  +  2x^  +  2x  +  l  =  Remainder 


As  above,  we  may  now  take,  instead  of  the  given  problem,  the  new 
problem  of  finding  the  H.  C.  F.  of  the  divisor  and  the  remainder : 

x3  +  2  x2  +  2  X  +  1 


x  +  2 


a;4  +  4  xs  +  2  x2  +  X  -  2 
a;4  ^  2  x«  +  2  x^  +     X 

2  x3  -  2 

2x8+4x2  +  4x+2 

—  4x2_4x  —  4  =  —  4  (x'^  +  X  +  1)  =  Remainder 

Again,  take  the  divisor  and  the  remainder  and  discard  the  factor  —  4, 
which  clearly  cannot  be  a  common  factor : 


x8  +  2x2  +  2x  +  l 

X2  +  X  +  1 

a;8  +     x^+     X 

x+1 

x'^+     X  +  1 

X2  +       X  +  1 

0 

=  Remainder 

Since  this  remainder  is  zero,  x^  +  x  +  1  is  a  factor  of  x^  +  2  x^  +  2  x  +  1 ; 
hence,  it  is  the  H.  C.  F.  of  the  given  expressions. 

At  any  stage  in  the  process  we  may  find  the  H.  C.  F.  by 
inspection,  as  in  example  1,  if  it  is  easy  to  do  so.  Other- 
wise we  repeat  the  same  process  as  often  as  is  necessary 
until  we  find  a  remainder  zero  ;  the  last  divisor  is  then 
the  required  H.  C.  F. 

Numerical  factors  may  he  inserted  or  taken  out  at  any 
time,  for  it  is  easy  to  see  by  inspection  whether  or  not  such 
factors  are  common  factors  of  the  given  expressions.  Usu- 
ally no  account  is  taken  of  these  purely  numerical  factors. 

Any  common  factors  that  can  be  found  by  inspection 


EUCLIDIAN   METHOD:    H.C.F.   AND   L.  C.  M.        375 

should  be  removed  at  once.  Sometimes  the  H.  C.  F.  can  be 
found  in  this  manner,  as  in  Chapter  V. 

The  L.  C.  M.  of  two  polynomials  can  be  found  by  §  74, 
p.  129,  after  finding  their  H.  C.  F. 

A  precisely  similar  process  holds  for  numbers  ;  but  care 
should  be  taken  not  to  insert  or  discard  numerical  factors 
in  dealing  with  problems  in  numbers. 

EXERCISES:    NOTE   VI  — H.C.F.    AND   L.C.M. 

[Solve  by  inspection  as  in  Chapter  V  wherever  possible;  otherwise 
use  the  process  explained  above.] 

\ 
Find  the  H.  C.  F.  and  the  L.  C.  M.  of : 

1.  35,  75,  25,  and  65. 

2.  42,  105,  147,  and  63. 

3.  1884  and  2079. 

4.  3718,  5269,  and  12,168. 

5.  35  a'b-,  84  a^b\  and  63  a^b\ 

6.  ic2_4^_2l  and  .i-^  — 5a;  — 14. 

7.  x^  —  1,  ar  —  1,  and  (x  —  1)-. 

8.  x^  +  3cr  +  Sx-\-2  and  x^  —  2x^  —  x  —  6. 

Find  the  H.  C.  F.  of : 

9.  af' ^_ ar' +  3 x  + 10  and  .>y  +  x-^- 5.T-6. 

10.  ar'  —  X*  —  x+1  and  5  .r*  —  4  ar*  —  1. 

11.  ar^  +  6  a:  +  9  and  .i"  +  .ar  +  x  +  21. 

12.  X*  -  2  .r'  +  3  X-  -  4  .T  +  2  and  .r'  -  2  a;*  +  3  x^  +  a^  -  8  a;  +  5. 

13.  4  a;^  - 18  a-  -  3  + 19  X  and  19  a,-^  -  12  ar*  +  2  .a;*  +  9  -  6  x. 

14.  a.-*  —  x^  +  x  —  1  and  a;^  +  af'  —  a;  —  1. 

15.  3  X*  -  4  X-  -  4  and  3  a;^  -  8  .r  +  4. 


376  APPENDIX 

16.  ar^  -  1  and  x^  +  2  or^  +  3  a;2  +  2  a;  +  1. 

17.  df  +  3x^  +  7  x  +  5,x*-2x^-\-4:X-  +  2x  —  5,  and  x^  +  5x^ 
\-x'  +  5. 

18.  ar^  -  5  a;  +  4,  a;*  -  2  a^  4- 1,  and  a;«  +  4  a^  -  3  a;  -  2. 

x^  +  Sx^  +  Ax^-dx'-Bx  +  G, 


"■    V  +  3a;^  +  7a^  +  10a;-12. 

(  4  +  7  x  +  x^  +  x^  -X*, 
^°-    |7a;  +  4-2a,'^-3ar'-3a;^ 

ja^-3ar'  +  a;<  +  3a;-2, 
^^'    |2a;-9ar'  +  4ar^  +  3. 

22.  a^-6a;2  +  lla;-6,    a;^  -  9  a;'^  +  26  a;  -  24,    and   a^-8a^ 
f  19  a; -12. 

Find  the  L.  C.  M.  of  the  following  : 

23.  9  xy-,  6  x^]f,  and  15  yh^. 

24.  a^—  1  and  ar^^- 1. 

25.  4  a;-  -  9  ?/'  and  4  a^  - 12  xy  +  9  y^. 

26.  (a  +  3),  (a^  -  9),  3  a  + 15,  and  15  a  -  45. 

27.  2  a^  -  a-  +  2  -  3  a;,  6  a^  +  4  .r'  -  4  -  2  a;,  and  4  x3  _  5  ^  _,_2, 

28.  a^  —  11  a;  +  24,  a;^  —  6  X  —  16,  and  a*^  —  a;  —  6. 

29.  c-  —  (a  +  &)-,  6-  —  (a  +  c)-,  and  a'  —  (b  +  cf. 

30.  (a;2  -  2  a.-  +  1),  (x'  -  l)^  a;=^  -  1. 

31.  :e*  —  .^•^  +  X  —  1  and  x'^  -\-a?  —x  —  \. 

32.  .r^  -  9  a.-2  +  26  a;  -  24  and  x'  -  10  .r  +  31  a;  -  30. 

33.  .r  +  6  a;  +  9  and  x?  +  .r-'  +  a;  +  21. 

34.  (a  +  hf  -  (c  +  df  and  (a  +  ?>)3  -  (c  +  d)='. 

35.  x'^  +  11  a;-  +  25  a;  and  x'  +  2  .r'^  +  11  ;c2  +  10  a-  +  25. 

36.  ar'  4-  3  a:^  ^  7  ^.  _^  5  ^nd  ar''  +  5  x^  +  x-  +  5. 

37.  a^  +  2/^,  a^  —  .v'',  (a;  —  y)^,  and  a^  +  3  x'^y  -\-  3  xy-  +  ?/l 

38.  7?  -\-2  xy  -\-  y'^,  x"^  —  2  xy  -[-  y^,  x?  —  y-,  and  x^  +  i/l 


NOTE  VII.  CUBE  ROOT  AND  HIGHER  ROOTS 


20.  Introduction.  When  cube  roots  and  higher  roots 
of  numbers  are  needed  in  practical  work,  they  can  be 
found  approximately  most  easily  by  the  use  of  loga- 
rithms (see  §  163),  which  computers  almost  always  use. 

Any  root  of  any  number  can  be  found  approximately  by 
trial,  as  in  §  97.     This  process  may  be  very  long. 

An  old  method,  analogous  to  that  used  in  §  97  for  square 
root,  is  given  here  chiefly  for  its  historic  interest. 

21.  Cube  Roots  of  Numbers. 

Ex.  1.    Find  the  cube  root  of  91609.86. 

Notice  that  l^  =  1,  10^  =  1000,  100^  =  1,000,000,  and  so  on.  Also, 
.P  =  .001,  .018  =  .000,001,  etc.  Mark  off  the  number  into  periods  of 
three  figures  each,  in  each  direction  from  the  decimal  point,  thus  : 
9i,609.86.     This  assists  in  estimating  the  first  figure  of  the  root. 

The  process  is  based  on  tlie  formula 

(a  +  by  =  «3  +  3  a%  +  3  uV^  +  b^  =  a^  -\-  Z»(3  a^  +  3ab  +  ?/-), 

and  consists  in  choosing  convenient  values  of  a  and  of  b  and  veri- 
fying ;  and  then  in  repeating  the  process,  choosing  as  a  new  value 
for  a  the  whole  root  already  found,  after  the  manner  of  §  97. 


408  = 
3  X  402  ^  4800 


91,609.86 

64,000 

27,609.86 

91,609.86 

91,125 

40 

5 

45 

45 

.07 

484.86 
91,609.86 
91,550.911843 

45.07 
45.07 

58.948157 

458  = 
3  X  452  =  6075 

45.078  = 


We  take  for  a  the  largest  convenient  number  whose  cube  is  con- 
tained in  the  given  number,  in  this  case  a  —  40.  If  b  is  the  remain- 
ing part  of  the  root,  tlien  3  a-b  +  3  ab'^  +  b^  is  the  remainder  of  tlie  cube. 
Since  b  is  small  compared  with  a,  the  principal  term  of  this  is  3  a-6. 

377 


378 


APPENDIX 


Hence,  3  a'^h  is  nearly  equal  to  27,609.86.  That  is,  3  x  402  x  &  = 
27,609.86  nearly.  Hence,  b  =  (27,609.86)  ^  (3  x  402)  nearly  =  5  nearly ; 
this  is  called  the  trial  divisor.  Hence,  the  root  is  a  little  over  45.  Start 
again  :  a  =  45,  cube  a,  subtract  a^  determine  h  =  .07,  etc.,  repeating 
this  process,  each  time  taking  as  a  the  part  of  the  root  already  found. 
Each  time  h  is  found  by  dividing  the  remainder  by  3  a2.  The  last 
two  figures  of  the  root  are  found  by  simple  division. 

This  work  can  be  tabulated  in  slightly  different  form,  as  follows  : 


91,609.86 

40 

403  = 

64,000 

3  X  402  =  4800 

27,609.86 

5 

3  X  40  X  5  =  600 

52  =  h 

5  X  5425  = 

27,125 

45 

3  X  452     =  6075 

484.86 

.07 

3  X  45  X  .07  =   9.45 

.072  =    .0049 

.07  X  6084.4549 

=   425.911843 

45.07 

3  X  45.072  =  6093.9147 

58.948157 

.0097 
45.0797 : 

Ansvier  to  four  places 

EXERCISES   I:   NOTE  VII  — CUBE   ROOTS   OF   NUMBERS 

Find  the  cube  root  of : 

1.  262,144.  3.    .001906624.  5.    12.812904. 

2.  69,426,531.  4.   259,694.072.  6.   64,144.108027. 

Find  the  cube  root  of  the  following  to  four  figures : 
7.   25,473.       8.  46.32.       9.  3.4674.       10.  65,463,257.0423. 

22.  Cube  Roots  of  Polynomials.  The  cube  root  of  a 
polyuouual  that  is  a  perfect  cube  may  be  found  in  a 
simihir  way.  The  polynomial  should  always  be  arranged 
according  to  the  ascending  or  descending  powers  of  the 
same  letter.  The  most  important  points  to  remember  are 
(1)  the  trial  divisor  is  3  times  the  square  of  the  part  of 
the  root  already  found ;  (2)  the  complete  divisor  is  3  a^  + 
^ah-\-h^  where  a  is  the  part  of  the  root  previously  found 
and  h  is  the  new  term. 


CUBE    ROOT    AND    HIGHER   ROOTS 


379 


Ex.    1.    Find  the  cube  root  of: 

114  X*  -  ill  a?  - 1.35  X  -  27  +  8  ./•  +  55  x"  -  60  x\ 

18x6-60x5  +  114x-»+    55x»- 171x2- 135X-27 

ISXB 


12  X* 

-30x3 

+  25x2 

-60x5 

-60x5 

+  114x<+    55x3 
+  150  X*  -  125  x3 

-171x2- 135  X- 27 

12  X* 

-30x^^  +  25x2 

12  X* 

-60x3  +  75x2 
-18x2 

+  45x 

+  y 

-  36  X*  +  180  x3 

-  36  X*  +  180  x3 

-171x2-135x-27 

12  X*- 

-60x3  +  57x2 

+  45  X  +  9 

-  171  x2  -  135  X  -  27 

2x2 


— 5x 


Answer :  |  2  x2  —  5x  —  3 

The  first  term  of  the  root  is  evadeiitly  2  x2 ;  hence,  the  first  trial  di- 
visor is  3(2  x2)  =  12  X*.  Dividing  the  first  term  of  the  first  remainder 
by  this  trial  divisor,  we  find  the  next  term  of  the  root :  —  5  x.  The 
first  divisor  completed  is,  therefore,  12  x*  +  {3(2  x2)(  — 5  x)  +  (  —  5  x)2}=: 
12  X*  —  30  x3  +  25  x2.     The  remaining  steps  are  repetitions  of  these. 

23.  Higher  Roots.  Since  the  fourth  power  is  the  square 
of  the  square,  the  fourth  root  is  the  square  root  of  the 
square  root.  Likewise,  the  sixth  root  is  the  cube  root  of 
the  square  root.  Other  roots  may  be  found  in  a  similar 
manner,  or  by  a  metliod  similar  to  that  used  for  cube  root. 
Any  root  of  any  number  can  be  found  approximately  by 
a  figure  (p.  190),  or  by  logarithms  (§  165,  p.  342). 

EXERCISES   II:   NOTE   VII  — CUBE   ROOTS   OF   POLYNOMIALS 

Fiud  the  cube  root  of: 

1.  ^A3a?-Ula-b  +  \%^ab--21h^. 

2.  8  m""  -  12  m'n  +-  30  m'n^  -  25  nv'n^  +  30  mhi*  -  12  mn'  +  8  n\ 

3.  fc"  + 12  A;-'  +  63  k*  +  184 1^  +  315  k""  +  300  fc  + 125. 

4.  a^_12a^  +  54a-112  +  l^-^|  +  -^. 

a        a-      a^ 

5.  Qx*  +  ^^+\.  +  20  +  -,+x'+15x^. 

X'*         X  XT 

6.  184  c-''  +-  315  r  +-  63  c"  +-  300  c  + 125  + 12  c'  +  c  . 


NOTE    VIIL     LIMITS   AND    INFINITE   SERIES; 
IRRATIONAL   NUMBERS 

24.  Introduction.  In  this  article  we  shall  state  briefly 
the  fundamental  notions  connected  with  limits.  These 
propositions  are  of  a  far  more  intricate  character  than  are 
most  of  the  topics  treated  in  this  book,  and  it  is  recom- 
mended that  these  articles  be  studied  only  by  mature 
students. 

25.  Limits.     In  case  the  difference   hetiveen   a   variable 
quantify  v  and  a  constant  quantity  k  cayi  he  made  to  become      \ 
and  remain  as  small  numerically  as  ice  please^  the  variable  v  is 
said  to  approach  the_  constant  k  as  its  limit,  and  we  write 

Lim  V  =  k. 

Usually  we  shall  have  some  definite  process  for  deter- 
mininsf  whether  or  not  the  difference  between  the  variable 
and  the  constant  can  be  made  to  become  and  remain  as 
small  as  we  please. 

Ex.  1,    Imagine   any  porons    body  (e.g.  a  brick)  soaked  in 

water  till  its  weight  is  increased.     Let  k  be  the  weight  of  the 

body  when  dry,  and  let  v  be  the  weight  when  wet.     If  the  body 

is  heated,  the  water  passes  off  in  steam,  and  the  weight  v  is 

variable.     It  is  evident  that  by  continued  heating  we  can  drive 

off  as  much  of  the  water  as  we  please.     Hence,  we  can  make 

the  difference  v  —  k  become  and  remain  as  small  as  we  please ; 

we  say : 

Lim  V  =  k. 

Ex.  2.  It  is  found  by  trial  tliat  a  rubber  balloon  will  burst 
when  it  is  so  filled  with  gas  as  to  have  a  diameter  of  six  feet. 
If  V  means  the  volume  of  the  balloon,  it  is  evident  that  we  may     | 

380  I 


LIMITS   AND    INFINITE   SERIES  38i 

expand  balloon  so  that  its  volume  is  as  near  as  we  please  to 

i^  •  {6f  cu.  ft.  =  288  TT  cu.  ft. ;  hence, 
o 

Lim  V  =  288  tt. 

In  this  example  the  variable  t;  cannot  be  made  equal  to  its  limit, 
for  the  balloon  would  burst  if  expanded  to  exactly  six  feet  diameter. 

In  example  1,  on  the  contrary,  the  variable  may  equal  its  limit,  for 
we  may  drive  off  all  the  water  and  leave  v  =  k. 

Ex.  3.    Consider  the  expression  : 

1  -  CD" 

1-1 

We  may.  make  this  expression  become  and  remain  as  near  to       ^ 
as  we  please,  by  making  n  sufficiently  large. 


For  let  ^-11 

1  —  -^ 


1      _o 


Then,       ._,.i_a__A_  =  _xo_  =  _y   , 

;  1  1 

or,  w  —  « 


2"-l  2  .  2  •••  (^M-l)  times)-..  2 

Since  the  denominator  contains  as  many  factors  2  as  we  please,  this 
fraction  can  evidently  be  made  to  become  and  remain  as  small  as  we 
please,  numerically,  by  taking  n  large  enough.     Hence,  * 

where  we  understand  that  the  limit  is  to  be  taken  by  making  n  as 
large  as  we  please. 

Ex.  4.    Similarly,  1-fiX 


V 


where  x  is  any  number  numerically  greater  than  1,  approaches  as  its 


limit 


1  X 

or 


X 


382  APPENDIX 

For  let 


k^ 


1  _i~^-l' 


t;-/t-  = 


X  X  a; 

1 


(x  —  1)  •  X  •  X  •••  ({n-\)  times)  •••  X 

If  a;  is  numerically  greater  than  1,  the  denominator  can  be  made 
as  large  numerically  as  we  please ;  hence,  v  —  k  can  be  made  to  be- 
come and  remain  as  small  numerically  as  we  please  by  taking  re  suffi- 
ciently large.     Hence, 

T 


Lim 


1 


1-1 


X      J 
if  X  is  numerically  greater  than  1 


1  -  - 

X 


l~x-l' 


26.  Infinite  Series.  The  last  examples  above  have  a 
direct  application  in  the  question  of  infinite  series.  By  an 
infinite  series  we  mean  an  unending  sequence  of  terms 
connected  by  +  signs: 

^0  +  ^1  +  «2  +^3  +    ■*■   +  ^ra  +   •••• 

(It  is  improper  to  say  that  an  infinite  series  is  the  sum  of 
an  unending  sequence  of  terms,  for  the  word  "  sum  "  is 
defined  only  for  a  finite  number  of  terms.) 

We  cannot  find  the  sum  of  such  an  infinite  series 
directly,  but  we  can  add  together  as  many  of  the  terms  as 
we  please  ;  thus,  we  can  find  the  sum  of  the  first  n  terms  : 

'S'n  =  «o  +  ^1  +  *2  +    •  *  •   +  ^«-l  " 

by  direct  addition.  We  say  that  the  sum  of  the  series 
is  the  limit  of  S^  as  n  increases  indefinitely,  if  there  is  a 
limit :  g^  Lim  *S'„,     if  Lim  S^  exists. 


LIMITS   AND   INFINITE   SERIES 


383 


27.    Infinite  Geometric  Series.     As  an  example,  consider 
an  infinite  gcdinetric  seiic's  (see  §  159,  p.  328): 

a  +  ar  +  ar^  +  ar'^  +  ••  •  +  ar'^  +  •••, 

where  r  is  numerically  less  than  1.     The  sum  of  the  first 
n  terms  (see  p.  32!  >)  is 

\  fn 

Sn=  a+  ar  +  cir^  +  •  •  •  +  ar"~^  =  a  - 

1  — r 

The  sum  of  the  infinite  series  is  therefore 

S—  Lim  (/S'„)  =  Lim(  a  ■ j  • 

Since  r  is  numerically  less  than  1,  let  us  put  r  =  - ;  then, 
X  is  numerically  greater  than  1,  and  we  have 


S—  Lim  (  a  •  i '—  )=  Lim 


i-i 


I 


1-1 


-,  by  example  4,  §  25. 


or, 

8=a- =  - 

1  —  /•       1  —  r 

Hence,  the  sum  of  an  infinite  geometric  series  in  which  r  is 

numerically  less  than  1,  is  - — —  • 

1  —  r 

Ex.  1.    In  the  infinite  series 

a=l,  r  =  ^;  hence,  S  = = r  =  2. 


1-r     1-i 

Ex.  2.    The  series 

1  _  1  4_  i  _  1  _L  _i J_  4.   1    _  . . . 

a  1 


gives 


a  =  1,  r  =  —  i;  hence,  S  = 


1-r     l-(-i)      3 


384  APPENDIX 

Ex.  3.    The  series 

gives         a  =  3,  r  =  i;  hence,  S  =  ^  =  ^  =  ^  =  3|. 
Ex.  4.    The  series 

3  _1_  _3_  _1_  _  3 I         3       4_  . . . 

a     ^      3      ^30^10 
1-r     1-J^      9       3' 


gives        a  =  3,  r  =  ^;  hence,  S  — 


Note  :  this  series  may  be  written  8.8333  •••  in  the  form  of  a  repeat- 
ing decimal,  and  we  may  write  J,"-  =  3.3333  ••• 

Ex.  5.    The  repeating  decimal  .27272727  •••,  which  is  often 
indicated  by  .27,  is  equivalent  to  the  geometric  series : 

_2J_  _)_       2  7 |_  2  7.         -I-  . . .  • 

1  0  0     "^  1  0000     "^  1  00  OTTOT    "^         J 

here     a  =  "  ,  r  =  y^^ ;  hence,  *S  =  ^—  =  — lO—  =  ^  =  ^ ; 
10¥'  100'  '  1  _  >•       1_     1  QQ       11' 

consequently,  ^3.  =  . 272727  ••..  -^      '      -^      loo      ^^      -^-^ 

Ex.  6.    The   decimal    5.743216216216216  •••   (often    written 
5.743216)  repeats  the  figures  216  forever ;  it  is  equivalent  to  the 

'''^''  '>16     ^6 

10«      109  ^ 

Taking  a  =  ^,  r  =  ^3,  we  find 

216 

216     216      ^_^     10«  216  24 


10«      10^  i__L      999000      111000' 

10^ 

hence,     5.743216216216  =  5  +  ^^  +  -^i_  =  5  +  ^^^^^  . 

1000      111000  111000 

In  general,  any  repeating  decimal  may  be  regarded  as  a 
terminating  decimal  +  a  certain  geometric  series ;  such  a  deci- 
mal may  therefore  be  evaluated  by  the  preceding  formula  for 
the  sum  of  an  infinite  geometric  series ;  the  result  is  always  a 
rational  fraction. 


LIMITS    AND    INFINITE    SERIES  385 

EXERCISES   I:    NOTE   VIII —  INFINITE   GEOMETRIC   SERIES 
Find  the  values  of  the  following  infinite  geometric  series: 

4        5     I     1  5  _r_    4  5     _i      1  3  5     I     . . . 

2      2       2 

2*  H"^H2~H3~'~"'*  5       _9_,3_9_17._  _8_1 

O        O-        or  ^-  -^        2  8^       TJ        1  2  8 

.  Find  the  values  of  the  following  repeating  decimals ;  check 
each  answer  by  long  division : 

6.  2.222.-..  9.  5.133333. ••.  12.  10.1010101010  .... 

7.  .7777.".         10.  42.716161616....     13.  26.308308308-... 

8.  .232323-...     11.   .0454545..-.  14.   83.83838383-... 

28.  Other  Infinite  Series.  We  have  seen  hoAv  to  find 
tlie  sum  of  an  inlinite  geometric  series  if  r  is  numerically 
less  than  1. 

In  general,  as  in  §  26,  we  say  that  the  sum  of  any  series 

(1)  «o  +  ^i  +  ''2-l !-«„+••• 

is  S=  Lim S^^  —  Lim  {n^  +  a^  +  a^-\-  •••  +  rt„_j) 

if  this  limit  exists. 

If  Lim  8^  exists^  the  series  (1)  is  called  convergent ; 
otherwise  (1)  is  called  divergent. 

Ex.1.  Thus,  l  +  2  +  4-f-84----  is  an  infinite  geometric 
series  which  diverges,  for 

Si  =  1,  S.,  ^3,  S^  =  7,  S4  =  15,  • .  •, 

^,.=  (1  +  2  +  4+..- +2"-^), 

and  S„  becomes  greater  than  any  quantity  one  may  name; 

hence,  Lim  S„  does  not  exist,  and  the  series  diverges. 


386  APPENDIX 

Ex.  2.    Another  divergent  geometric  series  is 

1  —  1-fl  —  1  +  1— !•••,  where  a  =  1  and  r=  —  1; 

here  >S'„  =  + 1  if  n  is  even,  and  S^  =  0  if  n  is  odd.  Tliere  is 
therefore  no  constant  which  S„  approaches  ;  hence,  Lim  (/S'„) 
does  not  exist,  and  the  series  diverges. 

Ex.  3.    A  series  which  is  not  geometric,  and  which  diverges, 
is  i  +  i  +  |  +  i+|  +  i  +  |  +  x  +  ^+.... 

This  series  diverges,  for  if  Ave  arrange  it  in  the  form 

i+i+a+i)+(i+i+i+i)+-, 

inclosing  the  next  eight  terms  ending  with  J^,  then  16  terms 
ending  with  gV,  etc.,  we  see  that  each  parenthesis  exceeds  ^ ;  hence, 

^,^  >  1  4- 1  4-  i  +  •  •  •  (as  many  times  as  we  please)  +  h 
It  follows  that  S„  approaches  no  limit;  hence  this  series 
diverges. 

EXERCISES   II:   NOTE  VIII  — OTHER  INFINITE  SERIES 

1.  Show  that  the  series  1+3  +  5  +  7  +  9+---  diverges. 

2.  Show  that  the  series  2  — 1+3  —  1+4  —  iH diverges. 

3.  Show  that  the  series  1+1-1  + i  +  l  +  |--l  +  yV+--- 
diverges. 

4.  Show  that  any  series   A^  +  A^  +  Ao-] will  diverge   if 

An  does  not  approach  zero  as  n  increases  indefinitely. 

5.  Show  that  if  a^-}- ai  +  CL,+  ■■'  is  a  convergent  series  of 

positive  terms,  and  if  6o4-  ^i  +  ^^H is  another  series  of  positive 

terms  such  that  b„  <  a„,  then  the  second  series  converges  also. 

29.  Irrational  Numbers.  We  have  already  considered 
many  irrational  numbers;  thus,  V2  is  irrational.  (See 
pp.  184,  186.)  In  fact,  if  the  positive  integer  a  is  not  the 
perfect  square  of  an  integer,  Va  is  irrational,  for  the  sup- 
position Va  =  — ,  where  7n  and  n  are  each  integers  without 
n  2 

a  common  factor,  leads  to  the  absurdity  a=— ,  i.e.  an 
integer  is  equal  to  a  fraction. 


LIMITS    AM)    IXITNITE    SERIES  3«T 

A  rational  number  is  the  quotient  of  two  integers  ;  an 
irrational  number  is  any  leal  number  that  is  not  rational. 
(See  pp.  18t3,  284.) 

In  case  of  V2  we  saw  that  we  could  get  a  rational 
number  whose  square  is  as  close  to  2  as  we  please,  and 
we  wrote 

(1.4)2<2<  (1.5)2, 

(1.41)2<  2  <  (1.42)2, 

(1.414)2<  2  <  (1.415)2. 

In  fact,  the  square  of  any  rational  number  is  either 
greater  than  2  or  less  than  2.  Thus,  the  rational  numbers 
are  divided  into  two  classes  according  as  their  squares 
are  greater  than  or  less  than  2.  This  division  into  two 
classes  practically  defines  V2,  as  above. 

Whenever,  in  such  a  fashion,  all  rational  numbers  are 
divided  into  two  classes  such  that  any  number  in  one  of 
the  classes  is  greater  than  any  number  in  the  other  class, 
we  say  that  this  division  defines  a  certain  irrational  number^ 
which  is  called  the  cut  number.  Thus,  V2  is  the  cut  num- 
ber which  separates  the  rational  numbers  whose  squares 
are  less  than  2  from  those  whose  squares  are  greater 
than  2.  We  may  also  think  of  V2,  for  example,  as  the 
limit  of  a  sequence  of  rational  numbers 

where  (2— a2„)  approaches  zero  as  n  increases  indefinitely. 
Thus,  V2  is  tlie  limit  of  the  sequence  of  the  numbers 

1.4,  1.41,  1.414,  1.4142,  .••, 
which  are  obtained  in  the  ordinary  square  root  process. 

30.  Operations  on  Irrationals.  Any  irrational  can  be 
expressed  by  means  of  rationals  to  any  degree  of  exactness 


388  APPENDIX 

required.  Thus,  V2  may  be  expressed,  by  means  of  one 
of  the  numbers,  1.4,  1.41,  1.414,  •••to  any  desired  number 
of  decimal  phices.  Similarly,  for  any  irrational,  we  may 
decide  upon  two  integers  between  which  it  lies,  then 
tenths  between  these,  and  so  on ;  eventually  the  irrational 
is  expressed  to  any  number  of  decimal  places  desired. 

If  two  irrationals  are  given,  say  V2  and  V3,  we  may 
express  each  of  them  to  any  number  of  decimal  places. 
Having  done  so,  we  may  add,  subtract,  multiply,  or  divide 
these  approximate  values  of  the  two  to  get  the  sum,  dif- 
ference, product,  or  quotient  of  the  approximate  values. 

We  now  define  the  sum  of  two  irrationals,  or  of  one 
rational  and  one  irrational,  to  be  the  limit  approached  by 
the  sum  of  the  approximations  as  the  number  of  true 
decimal  places  in  each  approximation  increases  indefi- 
nitely. Likewise,  the  difference,  product,  quotient,  etc.,  are 
defined  as  the  limits  of  the  corresponding  approximate 
values.  It  can  be  shown  that  by  these  definitions  tlie 
axioms  of  §  24,  p.  35,  remain  true. 

As  an  example  consider  V2  x  V8.  The  rationals 
that  define  V2  are  all  those  whose  squares  are  less  than 
2 ;  the  rationals  that  define  V3  are  all  those  whose  squares 
are  less  than  3.  If  we  multiply  these  approximate  rational 
values  together,  it  is  clear  that  the  approximate  products 
are  all  those  rational  numbers  whose  squares  are  less  than 
6.  It  follows  that  V2  •  V3  =  V6,  a  result  previously  ob- 
tained by  analogy,  without  strictly  logical  proof.  Simi- 
larly, Va  .  V?=  Vai6,  and,  in  general,  ^'a  ■  ^l  =  ^ah.  This 
is  the  rule  of  p.  286,  which  may  now  be  strictly  proved. 

The  operations  on  radical  quantities  and  on  other  irra- 
tionals used  above  should  finally  be  revised  in  the  manner 
just  shown  in  order  to  justify  them  completely  ;  but  it 
will  not  be  advisable  to  do  this  in  detail  in  this  book.  It 
should  be  noticed  that  the  definition  given  on  p.  183,  by 


LIMITS   AND   INFINITE   SERIES  389 

means  of  the  o-rapli  y  —  x^^  really  includes  the  notion  of 
approximation  just  used.  In  general,  for  elementary  pur- 
poses, a  geometrical  definition  of  irrational  numbers,  by 
the  graph,  mentioned  on  p.  186,  will  be  found  most 
suitable,  since  it  is  easy  to  follow  and  is  also  quite  in 
accordance  with  the  most  logical  definitions. 

EXERCISES   III  :    NOTE   VIII  —  OPERATIONS   ON   IRRATIONALS 

1.  Write  out  a  detiuitiou  of  v  2  as  a  cut  number. 

2.  Give  five  terms  of  a  sequence  of  increasing  numbers 
whose  limit  is  Vo. 

3.  Give  five  terms  of  a  sequence  of  increasing  numbers 
whose  limit  is  v5 ;  also  a  sequence  of  decreasing  numbers 
whose  limit  is  v'o. 

4.  Multiply  VS  by  V5;  show  the  result  is  a  cut  number 
which  expresses  the  product  of  the  cut  numbers  which  give 
•\/3  and  V«">,  respectively. 

5.  Add  the  increasing  sequences  of  numbers  which  ap- 
proach V3  and  V5,  respectively,  term  by  term  ;  define  the 
sum  V3  +  V5  by  this  means  as  a  cid  number. 

6.  Prove  Va  X  V^  =  -\Jah.       7.    Prove  Va-^  V6  =  \/-- 

8.  Show  that  the  graphs  of  y  =  x^  and  y  =  2  serve  to  define 
the  cut  number  V2. 

9.  Show  that  the  graphs  y  =  x-  and  x-  +  y=\.  serve  to 
define  the  cut  numbers  x=  ±  V|^. 

10.  Show  that  the  graphs  of  y  =  x",  y  =  -^*,  y  =  ^i  serve  to 
define  all  cube,  fourth,  fifth  roots,     (See  p.  190.) 

11.  Show  that  the  curve  x  =  10"  (see  p.  336)  defines  log  x 
as  a  cut  number  for  every  value  of  x. 

Hint.  Show  that  10"  can  be  really  found  for  every  rational  y, 
hence  log  x  can  be  found  for  values  of  x  as  near  any  given  value  of  x, 
as  we  please. 


NOTE  IX.  IMAGINARY  AND  COMPLEX 
NUMBERS 

31.  Introduction.  In  the  work  of  this  book  we  have 
used  the  following  classes  of  numbers : 

(1)  Positive  integers,  defined  by  counting  (see  p.  1). 

(2)  Positive  rational  fractions,  the  quotients  of  pairs  of 
integers  (see  pp.  1,  118,  186). 

(3)  Zero,  defined  by  the  equation  0  +  a  =  a  (see  p.  15). 

(4)  Negative  integers  and  fractions,  defined  by  the 
equation  a  +(—  a)=  0  (see  p.  14). 

(5)  Irrational  numbers,  defined  by  a  cut  (p.  387). 
These  five  classes  of  numbers  are  called  real  numbers. 

Just  as  mankind  has  found  it  desirable  to  invent  and 
use  these  various  kinds  of  numbers,  it  is  convenient  to 
invent  and  use  a  sixth  kind  of  number. 

With  real  numbers  alone  we  may  perform  any  addi- 
tions, subtractions,  multiplications,  divisions,  except  divi- 
sion by  zero.  We  may  also  raise  any  number  to  any 
integral  power,  and  we  may  extract  integral  roots  of  posi- 
tive numbers.  Indeed,  we  may  extract  any  odd  root  of 
a  fiegative  number ;  but  an  even  root  of  a  yiegative  number 
cannot  he  expressed  by  the  numbers  we  know  at  present. 

For  this  reason,  the  new  kind  of  numbers  arise  in  solv- 
ing even  very  simple  quadratic  equations.  (See  pp.  182, 
212.)  Thus,  x^  +  \  =  0  gives  x^  =  -\oyx=±  V^^,  but 
V—  1  is  an  even  root  of  a  negative  number,  and  cannot  be 
expressed  in  terms  of  the  numbers  of  the  five  classes  above. 

32.  Definitions.  We  shall  denote  the  new  number 
V  —  1  by  the  letter  i : 

(1)  V^^=z,    i-2  =  (V^T)2  =  -l; 

390 


IMA(UNARY   AND   COMPLEX   NUMBERS  391 

and  we  shall  work  with  this  new  number,  as  far  as  pos- 
sible, according  to  the  rules  for  real  numbers,  observing 
always  the  axioms  of  §  24,  p.  35.      (See  note,  §  36.) 

The  product  hi  of  any  veal  numbi'r  h  and  the  new  num- 
ber i  is  called  a  pure  imaginary  number.  The  square  of 
such  a  number  is 

(2)  ihiy=  {hi) (bi)  =  b  -i-b  -1  =  bbii  =  bH"^  =  b\-l)=-b'^ 

by  IV,  p.  35.     Likewise,  (—  ^0^=  —  *^-     Hence,  we  say 

(3)  V^62  =  ±  bi,    and    V^^  =  V-(±V^O^  =  ±  ^x  ■  i 

if  a;  >  0,  so  that  the  square  root  of  any  negative  number  —  x 
may  be  'written  as  the  pure  imayinary  ±  Vx  ■  i.  We  shall 
always  do  this  at  once  in  any  problem.* 

The  sum  (r/  -f  6i),  where  a  and  b  are  real  numbers,  is 
called  a  complex  number,  or  simply  an  imaginary  number. 

33.  Direct  Operations.  Addition,  subtraction,  and  nuil- 
tiplication  are  performed  by  working  as  if  i  were  an  un- 
known letter,  and  replacing  i"^  by  —  1  wherever  it  occurs. 

Ex.1.    (2  +  30  +  (4  +  5i)  =  6  +  8/, 
and,  in  general,  (a  +  hi)  +  {c  +  di)  =  (a  +  c)  +  (6  +  d)i. 

Ex.  2.    (2  +  3  0(4  +  5  /)  =8  +  22  i  + 15  r 

=8  +  22  /  - 15  =  -  7  +  22  i, 

and,  in  general,  (a  +  hi)(G  4-  di)  =  ac  +  (he  +  ad)i  -\-  hdi^ 

=  ac  +  (be  -f  ad)  i  —  hd 
=  (ac  —  bd)  +  (be  +  adyi. 

These  operations  employ  only  the  direct  results  of  the  axioms 
(p.  35),  and  the  statement  r  =  —1. 

*  There  is  not  the  same  advantage  in  dealing  with  square  roots  of  nega- 
tive quantities  and  with  square  roots  of  imaginary  numbers,  in  deciding 
which  of  the  two  answers  shall  be  denoted  by  the  sign  V  ,  as  there  was 
in  dealing  with  square  roots  of  positive  quantities.  In  examples,  however, 
we  shall  take  V—  x  —  +  Vx  i,  to  avoid  undesirable  complication. 


392  APPENDIX 

Ex.  3.    (2  +  3  i)(2  -30  =  2'- (3 1)'  =  4  -  (-  9)  =  13, 
and,  in  general, 

(a  +  bi){a  -  bi)  =  a-  -  (biy  ^a? -{-b'^=o? -\.b\ 

The  result  in  this  problem  does  not  contain  the  letter  «; 
hence,  it  is  a  real  number. 

34.  Conjugate  Complex  Numbers.  The  importance  of 
Ex.  3  above  leads  us  to  emphasize  it  by  calling  a—  hi  the 
conjugate  to  a  +  hi.  The  product  of  two  conjugate  complex 
numhers  a  +  hi  and  a—  hi  is  the  real  positive  number  cP-  +  6^. 
This  is  often  stated  by  saying  that  the  sum  of  two  squares 
is  the  product  of  the  imagina.ry  factors  a  +  hi  and  a  —  hi : 
(4)  a^  +  h'^  =  (a  +  hi)  (a  -  hi) . 

35.  Division.  Division  of  complex  numbers  is  based 
on  this  fact,  for  the  divisor  (or  denominator)  may  be 
made  real  by  multiplying  both  divisor  and  dividend  (or 
both  numerator  and  denominator)  by  the  conjugate  to  the 
divisor  (or  denominator). 

- 7  +  22  i ^  -  7  +  22  i      2-3^•^-14  +  65^•-66^:8 
^'    '      2-f  3i  2  +  '6i      '  2-3i  4-9 1^ 

__  14  +  65  i  +  66_52  +  65i^4_^g^.^ 


4  +  9  13 

Check :  (2  +  3  i)  (4  +  5  0  =  -  7  +  22  L     (See  Ex.  2,  §  33.) 

Ex.2.  i  =  l.z:i  =  zii^  =  r:i=-i. 

i      i      —I      —I-      +1 
(We  might  have  multiplied  numerator  and  denominator  by  +  i 
instead  of  —  i;  but  —  iwas  taken  in  order  to  observe  the  general  rule.) 

Ex.3.   _2_^^.l^-^2-2.-^,_. 
1  +  ?:    1  +  i     1-i        2 

4-5t     4-5i     3  +  2i     12~-li-10i^ 
Ex.  4.  —  — 


3_2i     3-2r     3  +  2i  9-4i2 

^  12-7^•  +  10 ^ 22 -li _ 23 _  ^^. 
9  +  4  13  13     13^* 


i 


IMAdlNARY   AND   COMTI.EX    XUiMHERS  393 

36.  Further  Examples.  If  square  roots  of  other  nega- 
tive numbers  than  —  1  occur,  we  reduce  them  at  once  to 
the  form  above  by  the  relation  (3),  i.e.   V—  x  =  ±^xi. 

In  the  following  examples  the  sign  +  is  chosen  in  each 
case : 

Ex.1.   (2  +  V=^)  +  (3  +  2V^ 

=  (2  +  2V^-f(3  +  2.3V^:o[)=5  +  8i\ 

Ex.  2.   (4- V^9)  X  (2  + V^ni3)  =  (4  -3  0(2  +  4  i) 

=  20  + 10  i. 

Ex  3    l±V^^l+2j     1  +  2^^-3  +  4^^     3     4. 
*    ■  1  _  V^^     1  -  2  i      1  +  2  (•  5  5      5" 

Ex.  4.  (2  +  V^3)(l  +  V^  =  (2  +  V3  i)(l  +  V3  i) 

=  2  +  3  V3  /  +  3  /2=  _  1  +  3V3i. 

4+V35     4  +  V5i     2-V3i 
Ex.  5.    ; = 7^-  •  TT- 

2  +  V-3     2  +  V3i     2-V3i 

^  8  +  (2  V5  -  4  V3)  i  -  Vl5  P 

4:-3i' 

^(8  +  Vi5)  +  (2V5-4V3)^ 

7 

^  8  +  VI5     2V5-4V3^. 

7  7 

Failure  to  follow  the  directions  given  above  may  lead  to 
error,  for  although  some  of  the  rules  of  ordinary  algebra  hold, 
there  are  other  rules  which  do  not  hold  (under  the  agreements 
we  have  made)  for  imaginary  numbers.     For  example, 

V^  V^=l  =  ^■  •  i  =  |2  =  -  1 ; 

but  if   we  attempted  to  use   the    rule    of   ordinary  algebra: 
Va  V^  =  V«6,   we   should  obtain  the  incorrect  result 


V-iV-i=  +  V+i=i. 


394  APPENDIX 

for  we  have  agreed  in  this  book  that  V+l=-|-V+l  =  +  l. 
Avoid  the  rule  Vet  V6=  Va6  for  iinagiuary  numbers,  until 
after  very  much  more  thorough  study  of  the  subject,  by  pro- 
ceeding as  directed  above. 

EXERCISES   I:    NOTE  IX— OPERATIONS   ON   IMAGINARIES 

Perform  the  operations  indicated,  where  i  means  v^— 1; 
when  other  square  roots  of  negative  quantities  occur,  take 
V—  u;  =  +  Va;  i  in  these  exercises  : 

1.   (V^^)^-  13.  2V^^  +  5i-3V^^. 

2-   (-3*T-  14.   (2+ V^  +  (3-2V^. 

15.  (4  +  V^(4-V^. 

16.  (3  +  V^(4-V^. 
5-V^4 


3.   (2  0(3  0. 


4. 

(3  +  20  +  (2+50 

5. 

(6-40-(3+0- 

6. 

(1-0^ 

7. 

(4  +  50(3-20- 

a 

2-3^• 

o. 

2  +  3  i 

9. 

3 

2^•' 

10. 

6-i 

2-5i 

11. 

2  -  4  t  +      ^      . 
5-8i 

12. 

3V-4  +  7^•. 

17. 


18. 


19. 


2  +  V-4 

3+y^. 

2_V^2 
1  +  V^^ 


3  +  V^^ 

20.  (_|  +  iV:^)(-i-iV^. 

21.  (-l  +  iV^)«. 

22.  (-i-^V=^)l 

Write  the  real  and  imaginary  factors  of  the  following : 

23.  X'  +  y-.  25.    9x-  +  16?/l  27.    x'^  —  y*. 

24.  m-  +  n^.  26.    4  m~  +  w^.  28.    IG  if''  +  ?/"*. 

37.  Quadratic  Equations.  In  solving  quadratic  equa- 
tions we  found  just  such  numbers  as  those  above.  We 
may  now  use  these  answers  intelligently ;  hence,  we  may 
regard  them  as  answers.  Occasionally,  such  answers  may 
liave  some  actual  meaning  in  a  problem. 


IMAGINARY    AND   CUMi'l.KX   ^UMBERy  395 

Ex.1.  Given  .r^  +  2  =  0,  we  found  (p.  210)  x=±V^^. 
These  answers  actually  satisfy  the  equation,  for 

(±V^2)-+2  =  -2  +  2  =  0. 

Ex.2.  Given  .-ir  — -i  x -\- o  =  0,  we  found  x=  -\-'2  ±  V—  1, 
or  x  =  2  ±  i.     Substituting  these  answers  for  x,  we  find  : 

(1)  for  x  =  2  +  i: 

(2)  for  x  =  2-i: 

(2-/)2_4(2-0  +  5  =  (.3-4  0-(8-4  0+5  =  0; 

hence,  these  answers  really  satisfy  the  given  equation. 

Ex.  3.   Given  any  quadratic  aic^+  bx-\-c=0,  we  found  (p.  213) 

b    ,  V6^  — 4ac 

x  = ± . 

2  a  2  a 

The  answers  are  real,  equal,  or  imaginary,  according  as  b^  —  iac  is 
>0,  =  0,  <0.  See  p.  214.  The  only  difference  in  our  present  rules 
from  those  of  p.  214  is  that  we  now  understand  how  to  work  with 
imaginaries,  whereas  such  numbers  were  then  meaningless. 

EXERCISES   II:    NOTE   IX  — QUADRATIC   EQUATIONS 

Solve  the  following  quadratic  equations;  check  each  result: 

1.  ar-2a-  +  2  =  0.  5.    2  .r2-3  .e  + 5  =  0. 

2.  x^-4:X  +  8  =  0.  6.   ar-6.T  +  ll  =  0. 

3.  x'  +  Ax  +  8  =  0.  7.    a:2_3a,  +  3  =  0. 

4.  4a;2-4a-  +  10  =  0.  8.    x-  +  x+l  =  0. 

38.  Other  Operations.  Many  other  operations  in  imagi- 
naries are  p()ssil)le,  but  we  shall  content  ourselves  with 
the  preceding  after  giving  a  few  examples. 


Ex.1.     -^-l  =  ^V-l  =  yi  =  x  +  yi. 

Squaring  both  sides,  we  have     /  =  (x'^  —  ?/'-)  +  2  xyi, 
whence,  x'^  —  //'  =  0  and  2  t>/  =  1, 

or,  (x  —  y)(x  +  )/)  =  {)  and  2  xy  =  1. 


396  APPENDIX 

Solving  for  x  and  y,  we  find 

x^±-\\=±\^%    or,   x  =  ±^^=±iV2, 

but  the  last  result  may  be  discarded  since  we  wish  to  have  x  real. 

The  corresponding  values  of  y  are  y  —  ±  \^-- 

It  follows  that  x  +  yi=  ±1^/2  (\  —  i).     If   we   consider   also  the 
other  possibility    V—  1  =  V  —  i,  we  find  two  other  answers: 
x  +  yl^  ±i-^^(l  +  0- 

These  results  may  all  be  checked  by  raising  any  one  of  the  results 
to  the  fourth  power. 

Ex.  2.    Solve  the  equation  .x^  —  1  =  0. 

Noting  that  we  can  factor  the  left  side, 

(x  -  l)(x2+  X  +  1)  =0, 
we  see  that  a:  =  1,  or  else  a:^  +  x  +  1  =  0, 

an  equation  whose  roots  are  x  =  —  li^"^—  3. 

There  are  therefore  three  answers.     (Verify  each  of  them.) 

Ex.  3.    Solve  X*  +  x-  +  1  =  0.     Factoring,  we  find 

X"  +  x2  +  1   =    (X"  +  2  X2+   1)   _  x2  =    (x2  +   1)2  -  x^ 
=    [(X2  +   1)    -  X]  [(X2  +   1)    +  X] 
=   (X2-X+1)(X2  +  X  +   1). 

Hence,  x*  +  x^  +  1  =  0  gives  either 

x2  -  X  +  1  =  0,  or,     x2  +  X  +  1  =  0, 

hence,  x  =  ^  ±  I  V  —  3,  or,     x  =  —  ^  ±  |V—  3. 

Check  :  Substitute  x  =  |  +  ^  V—  3  in  x*  4  x^  +  1 ;  we  find 

G  +  W^3)*  +  Q  +  ^  v:ri)2  +  1  

=  (~  i_  -  W^rs)  +  (-.1  + iV31)+i=o. 

The  answers  x  =  J  —  2  V—  3,  x  =  —  ^  ±  ^V—  3  may  be  verified  in 
an  exactly  similar  way. 

No  other  problems  will  be  solved  because  the  best 
method  —  known  as  De  Moivre's  theorem  —  is  beyond 
the  scope  of  this  book.  Logarithms  also  lead  to  imagi- 
nary numbers ;  tlie  discussion  of  these  and  other  more 
intricate  matters  is  left  to  more  advanced  courses. 


NOTE   X.     SIMULTANEOUS    QUADRATICS 

39.  One  Equation  Factorable.  We  have  studied  in  Chap- 
ter X  pairs  of  simultaneous  equations  of  which  at  least 
one  is  a  quadratic  equation.  We  shall  add,  in  this  note, 
several  additional  methods. 

We  saw  that  any  such  pair,  one  of  which  is  linear,  may 
be  solved  by  substitution  (p.  255).  This  method  applies 
also  whenever  one  of  the  equations  can  be  factored  as  in 
the  following  examples. 

j,^  ^     {2x^-Zxy  +  f  =  0,  (1) 

U-2  +  /-10a;=75.  (2) 

(1)  may  be  written  in  the  form 

C2x-y){x-y)=0,  (1) 

which  is  equivalent  to  the  two  equations 

(la)     2a;-y  =  0;  (1  i)     x-y  =  Q. 

Solving  each  of  these  in  combination  with  (2)  by  the  method  of 
p.  2.56,  we  find 

from  (1  a)  :  from  (1  6)  : 

ly  =  10,J  U  =  -6.J  1;,  =  |+5V7,J  l,=  |-oV,.J 

These  answers  may  all  be  verified  by  the  student. 

These  solutions  are  shown  in  Fig.  74  by  the  points  marked  A, 
B,  C,  D,  respectively.  Equation  (2)  gives  the  circle  of  center  (.3,  0) 
and  radius  10;  equation  (1)  gives  the  two  straight  lines  (1  a)  and  (1  h). 

This  process  may  he  used  whenever  one  of  the  equations  is 
of  the  form 

(I)  Ax^  +  Bxy  +  Cy'^  =  0, 

for  this  equation  is  always  factorable  by  the  methods  of 
§§  61,  119,  pp.  98-225.  Such  an  equation  as  (I),  i.e.  an 
equation  in  which  the  degree  of  each  term  is  the  same,  is 
called  homogeneous. 

397 


APPENDIX 


Y 

1 

t 

1 

t 

/ 

1 

j' 

t 

j' 

_L 

j' 

t             Jl 

' 

/ 

irt               A 

) 

1 

/ 

1 

7 

A. 

I 

/ 

i 

^1 — ^j/i     1  1 

y 

1 

^r\ 

y 

zt    ^^ 

/ 

/ 

\(-2 

/ 

/ 

/ 

V 

J- 

/ 

/ 

^ 

i_ 

1 

/ 

/ 

x 

\ 

/ 

/ 

]T 

// 

0 

// 

v- 

y 

// 

/ 

■^ 

d 

/ 

'  /I 

t 

\^h  1 

J. 

/ 

JH\ 

7 

/ 

' 

/ 

\ 

Z 

/ 

^ 

/ 

\ 

^ 

' 

i 

s 

V. 

^^ 

^ 

1 

*^ 

/ 

1 

/ 

/ 

1 

A- 

Fig.  74. 


A  case  of  another  type  is  illustrated  by  the  example 
foUoAvinsf : 


Factoring  (1),  gives    {x  —  l/)(x  +  >/)  +  2(.r  +  t/)  =  0, 
or,  (x  +  y)(.r-~y  +  2)  =  0, 

which  is  equivalent  to  the  two  equations  : 

(la)     x  +  y  =  0;  {I  l>)     x-i/  +  2  =  0. 


(1) 
(2) 


SIMULTANEOUS   QUADRATICS 


399 


Solving  each  of  tliese  in  combination  with  (2)  by  the  methods  of 
p.  255,  gives 

from  (1  a):  from  (1  i)  : 

These  may  be  verified  by  the  student.     The  figure  for  (1)  in   Fig.  75 
is  a  pair  of  lines  (1  a)  and  (1  h)  ;  tliat  for  (2)  is   the  curve  y  -  x\ 


1 

\ 

/! 

r 

/ 

I 

/ 

i 

± 

r 

I 

i 

r 

/ 

/ 

^ 

/ 

i 

/ 

L 

/ 

/ 

±                  T 

/ 

/ 

^ 

'  ^ 

/ 

/ 

'J  (~, 

s:        1         b)  ' 

06] 

/ 

S                        Y~ 

^ 

i\                     \ 

y 

1      1    \  "  '              \ 

o                           Z 

^ 

"                  / 

I 

Z        i^L 

1 

z      o;^ 

iKlak           V 

X        1/ 

2^    4l      IL 

I           ^                ^ 

\           Zu                t 

X      7                   ^ 

^   z                  / 

(-1,  + 

^a^    1          L 

.^^              ^z 

^i 

_,(! 

-  \xi           t 

-7^- 

-  -K^        ^ 

1 

7 

-Z     - 

^^Sil^T 

1 

^      -7^     - 

i_               "  >^ 

7 

S 

i: 

-4=  ^^ 

3        -^ 

5 

s 

s 

1 

\, 

n:±    : 

—i 

S 

s 

Fig.  75. 


400 


APPENDIX 


which  we  have  drawn  before.  The  point  (-1),  (4- 1)  appears  twice 
as  a  sokxtion ;  the  figure  makes  clear  why  this  is  true. 

In  general,  any  quadratic  equation  containing  ^^  may  be 
factored  as  in  Example  2  if  an  attempted  solution  for  i/ 
involves  no  radicals  in  x. 

Ex.  3.    The  equation 
(1)  y2_Sx-  —  2xy-]-Sx  — 4.71  +  8  =  0  may  be  written 

y2  _  2(x  +  2)?/  =  3  x=^  -  8  it-  -  3. 

Solving  for  y  by  the  usual  method  (see  p.  206  ),  we  find 

y  =  (x  +  2)±C2x-l), 

i.e.  (la)     y  =  8  X  +  1 ;  or,  (lb)     y  =  -  x  +  S. 

Equation  (1)  therefore  represents  a  pair  of  straight  lines  (la)  and 
(1  b)  (Fig.  76)  ;  hence,  if  (1)  occurs  as  one  of  a  pair  of  simultane- 
ous quadratics,  the  pair  may  be  solved  as  above. 


1 

1 

1 

\ 

1 

^ 

1 

\ 

/(la 

) 

\1W 

1 

\ 

1 

\ 

\ 

f 

\ 

/ 

i 

r 

\ 

Oi 

\ 

-1 

p 

-, 

\ 

1 

0 

\ 

1 

\ 

1 

K 

j  - 

-5 

\ 

j 

\ 

1 

v. 

1 

J 

1 

1 

_ 

_ 

_ 

Fig.  76. 


SIMULTANEOUS   QUADRATICS  401 

40.  Type  Ax^  +  Bxi/  +  Cif  =  B.  If  neither  of  the  given 
pair  of  quadratics  is  factorable  in  the  sense  of  §  39,  it  is 
often  possible  to  form  a  new  equation  from  them  which  is 
factorable. 

Ex.1.    {-'  +  42^^  =  5,  (1) 

\xy^\.  <2) 

Multiplying  both  sides  of  (1)  by  —  1,  both  sides  of  (2)  by  5,  and 
adding,  we  find 
(;i)  x2  -  .5  xz/  +  4  </2  =  0, 

which  is  factorable  (see  §  89)  and  is  equivalent  to 

(3  a)     X  -  4  2/  =  0,  (o  /;)     x  -  y  =  0. 

Solving  each  of  these  with  (2),  we  find 

from  (3  a)  :  from  (3  h)  : 


i 

or 

'  .r  =  -  2,  ■ 

f^=!'l 

■  or 

r.  =  -i. 

[y=h 

\l/  =  ~hl 

U  =  i,j 

U  =  -i- 

These  answers  may  be  verified  by  the  student. 

This  process  may  be  used  whenever  both  equations  are 
of  the  form 

(II)  /lx^  +  Bxy-\-Cy^  =  D; 

the  rule  is  to  eliminate  the  constant  term,  as  above,  so  as  to 
get  an  equation  of  the  type  (I).* 

*  This  type  of  equations  also  yields  to  the  following  method : 
Let  y  =z  vx  in  both  equations  ;    equate  the  values  of  x-  in  the  resulting 
equations,  and  solve  for  v.    Thus,  in  the  preceding  example, 

r. 

(1)  becomes  a;'^  +  4  v'^x^  =  5,  or  x^  = 

1  +  4  tJ"^ 

(2)  becomes  vx^  —  I,       x-  =  -- 

V 

5  1 

Equating  the  values  of  x^ :    =  -,  or  4  r^  —  5  u  +  1  =0. 

I  +  iv^      V 

Hence,  v  =  I  or  1,  whence  x  =  ±  2  or  ±  1,  and  y  =  vx  =  ±  1  or  ±1, 
which,  when  pi-operly  paired,  give  the  pairs  of  solutions  found  above. 


402  APPENDIX 

41.    Symmetrical  Type.     Other  equations  yield   to  the 
method  of  §  40. 

Ex.        (2xy-x-y  +  5  =  0,  (1) 

[  a:i^  +  6 xy  +  y"^  —  4: X  —  4:y  —  5  =  0.  (2) 

Multiply  both  sides  of  (1)  by  2 ;  multiply  both  sides  of  (2)  by  —  1, 
then  add :  (3)  x^  +  2  xij  +  i/^ -  2  x  -2  y  -  15  =  0, 

or,  (x-\-yy^-2(x  +  ?j)-15-0, 

which  is  the  same  as     \_(x  +  ?/)  —  5]  [(x  +  y)  +  3]  =  0, 

and  is  therefore  equivalent  to  the  two  equations 

(3  a)     x  +  y-5  =  0,  (ib)     x+/y  +  3  =  0. 

Solving  each  of  these  with  (1),  we  find 

from  (3  o)  :  from  (3  //)  : 


(  X  =  5,]         Cx  =  0, 1  (  X  —  \, 


1 


{x=-i,] 


This  method  is  successful  whenever  both  equations  are 
symmetrical ;  i.e.  of  the  form 
(III)  Ax--\-Bxy  +  Ay^  +  Dx  +  Dy  +  F=0, 

so  that  an  interchange  of  x  and  i/  does  not  affect  the  equa- 
tion ;  the  rule  is  to  make  the  quadratic  part  of  the  new 
equation  a  perfect  square,  bi/  midttpl^i7ig  each  equation 
by  the  value  of  B  —  2  A  in  the  other  equation,  and  then  sub- 
tracting ;  this  is  practically  what  we  did  above.* 

42.    General  Method  of  Inspection.     It  can  be  shown  that 
any   pair   of    simultaneous    quadratic    equations    can    be 

*  This  type  (both  equations  symmetrical)  can  be  solved  also  by  mak- 
ing the  substitution  x  =  u  +  v,  y  =  ti  —  v  ;  in  the  example  above  this  gives 
from  (1)  2  ?<2  -  2  v2  _  2  ?<  +  5  =  0, 
from  (2)  8  7f2  _  4  ^2  _  8  M  -  5  =  0. 

Adding  these,  after  multiplying  the  first  by  —  2,  we  get 
4u-2  _  4„  _  15  =  0, 
which  gives  ?*  =  f  or  —  f  ;  hence,  v^  =  \  (S  ifi  —  8  r<  —  5)  =  -^^  and  v  =  ±  f, 
whence,  jc  =  ?t  +  ■o  =  5  or  0,  or  1  or  —  4,  and  ?/  =  i«  —  v  =  0  or  5,  or  —  4 
or  1,  which  give  the  preceding  solutions  when  properly  paired. 


SLMULTAXKOUS   QUADRATICS 


403 


solved  ill  a  similar  manner  if  the  proper  multipliers  can 
be  foujid.  Thus,  on  multiplying  both  sides  of  the  first 
by  X  and  adding  it  to  tlie  second,  tliere  are  always  three 
values  of  X  which  make  the  new  e([uation  factorable 
in  the  sense  of  §  39.  In  many  examples,  however,  it  may 
be  impossible  for  the  student  at  pi'esent  to  find  these  values 
of\. 

Ex   1    {•'^-•'^7/  +  r  +  2.r-^  =  3,  (1) 

■   l.t-  +  /  +  4x-2?/  =  5.  (2) 

Choose  A.  =  —  2 ;  i.e.  multiply  (1)  by  —  2  and  then  add  equation  (2), 

or,  clianging  signs  and  transposing,  we  have 
(x-^)-^-l  =  0, 

whence,  (3)  is  equivalent  to  the  two  equations 

(;j  „)     X  -  ^  +  1  =  0,  Qlh)     x-y-\=0. 


1    [X^  '     ■'--^  ■  '                1              1 

^00^                                                              ^*S1^        ■**                                     j                                               fN 

■     .^    "                      ""      M^^            ^X    --^^     ^t-    -1 ^ 

/                   s    ^\  ^-  -,^ 

y_                                       •">     k       1  /     /^ 

/                                         \^     \        f    y  ^ 

J                                           1        \     /  </ 

J                                                                                                                                                                                 1                                   \              /y/ 

/                                                                                                   L^ "^^Jv  //                                                ' 

^^       ^                 S^^ 

/                                                l-il   ^<^          1    *             /^'Vs 

r                     '    *                /'/iW                       '^V' 

7                                      P              y^/    1  \                                  y' 

n  J                         i      X    /     1                 / 

X                              \^   1       \                / 

y                 ^    /i          y      ,  -    ■ 

/                             7       /        \    1      /  ^'<^S' 

J                                                       y             r.    A          \          \1/  -^ 

\                                 /                              /        10  /]      '      )U^  1                             Ml 

^5  >             ■-■                11*3           -2  :    :    1  j/-1             1/1      ^-^ari,    1    1    1    J       1           ;l 

\               /            ^^y         /  ^^  //i\       ' 

^L                                            /                                        ^                              h^///\ 

\           \            y      \  ^r  ^x// 

^V                7                  7             >^/~\/-//    \ 

\                    ^       -<  1  /  /// 

Is                              /      .-f            '/X'/^^>    1 

s      A         Z%=^             '/y>    -X-\&^ 

iX'   y/               "^ " yV 

^■',        /                                                             /    / 

iiciV       /                            / 1 

Y      ^            i  rr  y^,/*         ^3                           -^ 

1      /                  !'>!•'!               1 

1  J'^                 A  K    !  /                                                     i 

^^             >^      y^                                   Mi 

l'!lji|                                       (^|l_i'                                 ~^                                                            iJ_'                     1 

FiQ.  77. 


404  APPENDIX 

Combining  each  of  these  with  (2),  we  find  the  soUitious  (Fig.  77) 
from  (3  a)  :  from  (3  b)  : 

[y=2,\        [y=-2,j  U  =  0,  J         [y  =  -2,j 

which  the  student  may  verify  directly.     The  figure  shows  the  picture 
for  each  of  the  equations  used,  and  also  those  which  follow. 

The  choice  =  —  1  is  equally  successful  (Fig.  77,  .5  a  and  5  b). 

The  choice  =  —  f  is  equally  successful  (Fig.  77,  4  a  and  4  b). 

As  a  general  method  this  is  undoubtedly  superior  ;  its 
details  are  too  lengthy  for  discussion  here.  Notice  all 
the  methods  given  above  are  special  cases  of  this  one. 

EXERCISES    I:    NOTE   X  —  SIMULTANEOUS    QUADRATICS 

Solve  each  of  the  following  pah-s  of  equations  and  draw  a 
figure  showing  the  curves  and  their  points  of  intersection  : 


■y-  =  0,  {x^  —  5xy  +  4:y^  =  0, 


1.  <  3.    , 

\x^-2y  =  0.  [xy  =  3Q. 

j  a-2  -  5  xy  +  6  7/2  =  0,  j  ar^  -  2  xy + y'  -  3(x  -  y)  =  0, 

\x^-\-y^=100.  '    \y  =  x'\ 

f-  _  3  x"  -  2  a;?/  +  8 ;«  -  4  ?/  +  '"  =  0  (see  p.  400), 
y  =  x'. 

I  x"  -f^  33,  1 .6--  +  xy  +  f  =  21 , 

\xy-\.y''  =  U.  '    \x-—xy  +  y--2x-2y  =  23. 

I  x"" + 3 xy +y-  +  2x+2y  =  lT,  |  x-  -  xy  +  y-  =  3, 


10. 


11. 


x-2  +  4  xy  +  ?/2  +  4  .c  +  4  //  =  89, 
x^-2xy  +  y'-2x-2y  =  n. 

x^  —  2  xy  +  y-  +  4  .r  +  4  ^  =  16, 
2x'  +  xy  +  2y'-2x-2y  =  7. 


SUMMARY  405 


SUMMARY   OF   APPENDIX,   pp.  354-404 

Note  I.    Detachkd  Coefficients.  pp.  354-355. 

1.  Detached  Coefficients.  p.  354. 

2.  Multiplication :  illustrative  examples.  p.  354. 

3.  Division :  illustrative  example.  pp.  3.")4-355. 

4.  Division  hy  (^x  —  a)  :  illustrative  example.  Exercises  I. 

pp.  355-356. 

Note  II.    Remaindek  Theorem  ;  Factohing.  pp.  357-360. 

5.  Factor  Theorem:  proof  of  theorem  ;  converse.  p.  357. 

6.  Factors  of  x"  ±  y'* :    application  of  §  5 ;     type-forms.       Exer- 
cises I.  pp.  357-359. 

7.  i^ac/ors  o/Po/^nwn/a/s:  factors  by  trial.     Exercises  II.    p.  359. 

8.  Remainder  Theorem :  extension  of  §  5 ;  calculation  of  polyno- 
mial values.     Exercises  III.  p.  360. 

Note  III.   Choice  and    Chance;   Permutations    and    Combina- 
tions, pp.  361-365. 

9.  Choice:  general  formula.  p.  361. 

10.  C/iance :  general  formula.     Exercises  I.  pp.  361-362. 

11.  Permutations:  definition;  general  formula.  p.  363. 

12.  Permutations  among   a   Limited  Number:  discussion;  fornmla. 

pp.  363-364. 

13.  Factorial   Notation:  definition    of    n\  Pn,n'i    Pn,m',   formulas. 
Exercises  II.  p.  364. 

14.  Combinations:  definition;  formula.     Exercises  HI. 

pp.  364-365. 

Note  IV.   Inequalities.  pp.  366-368. 

15.  Operations  on  inequalities.  jt.  366. 

16.  Graphical  solutions.     Exercises  I.  pp.  367-368. 

Note  V.    Binomial  Theorem.  pp.  369-372. 

17.  Formula :  proof  for  positive  integral  exponents ;  mathemati- 
cal induction.  pp.  369-.j71. 

18.  Notes  and  Examples.     Exercises  I.  pp.  371-372. 

Note  VI.   Euclidian  Method;  II.C.F.  and  L.C.M. 

pp.  373-37(5. 

19.  Euclidian  method  :  examples ;  statement  of  principles.     Exer- 
cises I.  pp.  373-376. 


406  APPENDIX 

Note  VII.   Cube  Root  and  Higher  Roots.  pp.  .377-379. 

20.  Introduction.  p.  377. 

21.  Cube  Roots  of  Numbers.     Exercises  I.  pp.  377-378. 

22.  Cube  Roots  of  Polynomials :  example.  pp.  378-379. 

23.  Higher  Roots.     P^xercises  II.  p.  379. 

Note  VIII.   Limits  and  Ixfinite  Series;  Irrational  Numbers. 

pp.  380-389. 

24.  Introduction.  p.  380. 

25.  Limits:  definition  ;  illustrations;  examples.  pp.  380-382. 

26.  Infinite  Series:  definition;  definition  of  sum.  p.  382. 

27.  Infinite  Geometric  Progressions :   general  formula  for  sum ;  ex- 
amples ;  repeating  decimals.     Exercises  I.  pp.  383-;38;j. 

28.  Other  Infinite   Series:   definitions  of  convergence,  divergence ; 
examples.     Exercises  IT.  pp.  385-386. 

29.  Irrational   Numbers:    definitions;    illustrations;   direct  defini- 
tions by  "  cuts."  pp.  386-387. 

30.  Operations     on    Irrationals:     approximations;     definition     of 
"sum,"  etc.     Exercises  III.  pp.  387-389. 

Note  IX.   Imaginary  and  Complex  Numbers.  pp.  390-396. 

31.  Introduction.  p.  390. 

32.  Definitions.  pp.  390-391. 
38.  Direct  Operations.  pp.  391-392. 
84.    Conjugate  Complex  Numbers.  p.  392. 

35.  Division.  p.  392. 

36.  Further  Examples.     Exercises  I.  pp.  .393-394. 

37.  Quadratic  Equations :  imaginary  solutions.     Exercises  II. 

pp.  394-395. 

38.  Other  Operations  :  examples.  pp.  395-396. 

Note  X.   Simultaneous  Quadratics.  pp.  397-404. 

89.    One  Equation  Factorable  :  homogeneous  type,  A  x^  +  Bxn  +  Cy^. 

pp.  397-400. 

40.  Type  Ax^  +  Bxy  +  Ci/  =  D:  elimination  of  constant,     p.  401. 

41.  Symmetrical  Type  :  Ax'^  +  Bxy  +  Ay^  +  Dx  +  Dy  +  F  =  0. 

p.  402. 

42.  General  Method  of  Inspection:  three  possible  multipliers.     Ex- 
ercises I.  pp.  402-404. 


TABLES 

[The  student  doubtless  knows  many  of  these ;  some  others  should  be  learned, 
stress  being  laid  ou  the  tables  in  the  metric  system.] 

I.   TABLE  OF  SIGNS 

+  ,  re&d  plus.     — ,  read  77ihiits. 

X  ,  or  •,  read  times,  multiplied  by,  multiplied  into,  or  into  (ay.h  =  a  •  b  =  ah). 
H-,  : ,  or  /,  read  dicided  by  or  over. 

=  ,  read  equals,  or  is  equal  to. 

^,  read  is  not  equal  to. 

>,  read  is  greater  than.  >,  read  is  greater  than  or  equal  to. 

<,  read  is  less  than.  <,  read  is  less  than  or  equal  to. 

[The  signs  just  written  may  be  easily  remembered  by  noting  that  the 
opening  faces  the  greater  quantity.] 

a^,  read  a  square  =  a  x  a.  a^,  read  a  cube  =  a  x  a  x  a. 

a",  read  a  to  the  nth  poiL'er,  a  to  the  jtotcer  n,  or  a  tcith  an  exponent 
n  =  a  X  a  •••  to  7j  factors. 

(  )>    [  ]>   {  }>       '~f  called  signs  of  aggregation;  a  general  term 
used  for  them  is  parentheses,  and  quantities  inclosed  by  them 
are  read  the  quantity  ■■■  or    the   expression  •■■.     To   distinguish 
them,  we  call  (  )  parentheses ;   [  ]  brackets,  {  }  braces ; 
the  vinculum. 

Va,  read  the  square  root  of  a ;  this  is  also  written  a-  and  read  a  to 
the  power  J,  see  pp.  192,  285. 

v'tt,  read  the  cube  root  of  a;  this  is  also  written  as  and  read   a   to 
the  power  j,  see  pp.  192,  285. 

va,  read  the  nth  root  of  a :  this  is  also  written  a>',  and   read    a   to 
the  power  i,  see  pp.  192,  285. 
407 


408 


TABLES 


II.    TABLES   OF   WEIGHTS   AND   MEASURES 

[The  customary  abbreviation  of  each  measure  is  indicated.     The  most  im- 
portant units  are  printed  in  black-faced  type.] 

A.    Measures  of  Quantity 


1.    Table    of  Measures  of  Common 
Objects 

12  units  =  1  dozen  (doz.). 

20  units  =  1  score. 

12  dozen  =  1  gross  (gr.)- 

12  gross   =  1  great  gross  (G.  gr.). 


2.    Table  of  Measures  of 

Paper 

24  sheets     =  1  quire. 

20  quires     =  1  ream. 

2  reams      =  1  htmdle. 

5  bundles  =  1  bale. 


B.   Measures  of  Value 


].    Table  of  American  Money 
10  cents  (^)  or  (ct.)  =  1  dime. 


10  dimes 
10  dollars 


1  dollar 
1  eagle. 


Other    measures  are    the    quarter 


2.    Table  of  English  Money 
4  farthings  (far.)  =  1  penny  (d.). 
12  pence  =  1  shilling  (>-.). 

20  shillings  =  1  pound  (£). 

Other  measures  are  the  crown  (=  5 


(=25  cents),    tlie    half-dollar   (=50    shillings)  and  the  guinea   (=21  shil- 
cents),  and  the  mill  (=x'5  cent). 


3.    Table  of  French  Money 
10  millimes  =  1  centime. 
10  centimes  =  1  decime. 
10  decimes    =  1  franc  (fr.). 


lings). 

4.    Table  of  German  Money 
100  pfennigs  (pf.)  =  1  mark  (nik.). 

5.    Table  of  Italian  Money 
100  centisimos  =  1  lira  (Ir.). 

In  the  following  tables  of  equivalents,  three  significant  Ji(/ures  are 
given,  except  in  a  few  important  instances.  The  student  should 
realize  that  it  is  not  the  number  of  decimal  places,  but  rather  the  num- 
ber of  significant  figures,  which  determines  the  degree  of  accuracy. 

6.    Table  of  Equivalents 
hi.    =.$0.0203.  1  fr.    =  SB  0.193. 

Is.     =$0,243.  lmk.  =  f  0.238. 

l£   =$4.8665.  1  Ir.     =$0,193. 

C.  Measures  of  Length 
1.    Table  of  English  Measure  of  Length 

12  inches  (in.)=  1  foot  (ft.). 
3  feet  =  1  yard  (yd.). 

5^  yards  =  1  rod  (rd.). 

320  rods  =  1  mile  (mi.). 


WEIGHTS    AND   MEASURES  409 

2.    Table  for  Metric  Measure  of  Length 

10  millimeters  (mm.)=  1  centimeter  (cm.). 

10  centimeters  =  1  decimeter  (dm.). 

10  decimeters  =  1  meter  (m.). 

10  meters  =  1  dekameter  (Dm.). 

10  dekameters  =  1  hektometer  (Hm.). 

10  hektometers  =  1  kilometer  (Km.). 

10  kilometers  =  1  myriameter  (Mm.). 

The  meter  is  approximately  one  four-millionth  of  the  circumference 
of  the  earth,  measured  along  a  meridian  through  Dunkirk,  France. 

3.    Table  of  Equivalents 

ENGLISH    TO    METRIC  METRIC    TO    ENGLISH 

1  in.  =2.54  cm.  1cm.    =0.3937  in. 

1  ft.  =  30.5  cm.  1  m.     =  39.1  in.  =  3.28  ft. 

1  yd.  =  91.4  cm.  =  .914  m.  1  Km.  =  0.621  mi. 

1  mi.  =  1.61  Km. 

D.   Measures  of  Area 

1.  Table  of  English  Measure  of  Area 

144  square  inches  (sq.  in.)  =  l  square  foot  (sq.  ft.). 

9  square  feet  =  1  square  yard  (sq.  yd.). 

30 J  square  yards  =  1  square  rod  (sq.  rd.). 

160  square  rods  =  1  acre  (A.). 

640  acres  =  1  square  mile  (sq.  mi.). 

2.  Table  of  Metric  Measures  of  Area 

100  square  millimeters  (sq.  mi.)=  1  square  centimeter  (sq.  cm.). 
100  square  centimeters  =1  square  decimeter  (sq.  dm.). 

100  square  decimeters  =  1  square  meter. 

100  square  meters  =  1  square  dekameter  (sq.  Dm.)  or 

are  (a.). 
100  square  dekameters  =  1  square   hektometer    (sq.  Hm.) 

or  hektare. 
100  square  hektometers  =1  square  kilometer  (sq.  Km.). 


410  TABLES 

3.    Table  of  Equivalents 

ENGLISH    TO    METRIC  METRIC    TO    ENGLISH 

1  sq.  ill.  =  6.45  sq.  cm.  1  sq.  cm.  —  .155  sq.  in. 

1  sq.  yd.  =  0.835  sq.  m.  1  sq.  m.    =  1.20  sq.  yd. 

1  sq.  rd.  =  .253  a.  1  a.  =  3.95  sq.  rd. 

E.   Measures  of  Volume 

1.    Table  of  English  Measure  of  Volume 

1728  cubic  inches  (cu.  in.)  —  1  cubic  foot  (cu.  ft.). 
27  cubic  feet  =  1  cubic  yard  (cu.  yd.). 

128  cubic  feet  =  1  cord  (of  wood)  (cd.). 

2.    Table  of  Metric  Meastirements  of  Volume 
1000  cubic  millimeters  (cu.  mm.)   =  1  cubic  centimeter  (cu.  cm.). 
1000  cubic  centimeters  =  1  cubic  decimeter  (cu.  dm.). 

1000  cubic  decimeters  =  1  cubic  meter  (cu.  m.)  or 

1  stere  (of  wood)  (st.). 

3.    Table  of  Equivalents 

ENGLISH    TO    METRIC  METRIC    TO    ENGLISH 

1  CU.  in.    =  16.4  cu.  cm.  1  cu.  cm.  =  0.0612  cu.  in. 

1  cu.  yd.  =  0.76  cu.  m.  1  cu.  m.    =  1.31  cu.  yd. 

1  cd.         =  3.6  St.  1  St.  =  0.28  cd. 

F.   Measures  of  Time 

60  seconds  (sec.)  =  1  minute  (miii.). 

60  minutes  =  1  hour  (hr.). 

24  hours  =  1  day  (da.). 

7  days  =  1  week  (wk.). 

14  days  =  1  fortnight. 

12  months  =  1  year  (yr.). 

365  days  =  1  year. 

366  days  =  1  leap  year. 
10  years  =  1  decade. 

100  years  =  1  century. 

The  number  of  days  in  the  different  months  vary.  February  has  28  days, 
except  for  leap  year  when  the  extra  day  is  added  to  it,  Riving  it  2<t.  Septem- 
ber, April,  June,  and  November  each  have  .%  days.  The  other  months  each 
have  31  days.  Every  year  divisible  by  four,  except  CQt\tiei\ni^l  years  not  di^ 
visible  by  400,  is  a  leap  year. 


WEIGHTS   AND  MEASURES  -111 

G.   Measures  of  Capacity 

1.  Tables  of  English  Measurement  of  Capacity 

DRY    MEASURE  LIQUID    MEASURE 

2  pints  (pt.)  =  1  quart  (qt.).  4  gills  (gi.)  =  1  pint  (pt.)- 

8  quarts  =  1  peck  (pk.).  2  pints  =  1  quart  (qt.). 

4  pecks  =  1  bushel  (bu.).  4  tpiarts        =  1  gallon  (gal.). 

Other  measures  are  the  barrel  (bbl.  =  311  gal.)  and  the  liogshead 
(lihd.  =  2  bbl.). 
The  gallon  contains  2:31  cu.  in. 

2.  Tables  of  Metric  Measurement  of  Capacity 
10  milliliters  (ml.)     =  1  centiliter  (cl.). 

10  centiliters  =  1  deciliter  (dl.). 

10  deciliters  =  1  liter  (1.). 

10  liters  =  1  dekaliter  (Dl.)- 

10  dekaliters  =  1  hectoliter  (Hi.)- 

10  hectoliters  =  1  kiloliter  (Kl.). 

The  liter  contains  1  cu.  dm.  or  61.02  cu.  in. 
3.    Table  of  Equivalents 

ENGLISH    TO    METRIC  METRIC    TO    ENGLISH 

1  dry  qt.      =  1.10  1.  1  I.     =  0.908  dry  qt. 

1  liquid  qt.  =  0.947  1.  11.=  1.0567  liquid  qt. 

1  bu.  =  35.2  L  1  HI.  =  2.84  bu. 

1  gal.  =  3.79  1.  1  I.     =  0.264  gal. 

H.   Measures  of  Weight 
1.    Table  of  Avoirdupois  Weight 

16  drams  (dr.).  =  1  ounce  (oz.). 

16  ounces  =  1  pound  (lb.). 

100  pounds  =  1  hundredweight  (cwt.). 

112  pounds  =  1  long  (or  English)  hundredweight. 

20  hundredweight  =  1  ton  (T). 

20  long  hundredweight  =  1  long  (or  English)  ton. 

7000  grains  =  1  pound. 


412  TABLES 

2.   Table  of  Troy  Weight  3.    Table  of  Apothecaries''  Weight 

24  grains  (gr.)     =  1  pennyweight  20  grains  (gr.)  =  1  scruple  (3). 

(pwt.).  3  scruples         =  1  dram  (3). 

20  pennyweights  =  1  ounce  (oz.).       8  drams  =  1  ounce  (  5  )• 

12  ounces  =  1  pound  (lb).     12  ounces  =  1  pound  (lb.). 

4.    Table  of  Metric  Pleasure  of  Weight 
10  milligrams  (mg.)  =  1  centigram  (eg.). 
10  centigrams  =  1  decigram  (dg.). 

10  decigrams  =  1  gram  (g.). 

10  grams  =  1  dekagram  (Dg.). 

10  dekagrams  =  1  hektograra  (Hg.). 

10  hektograms  =  1  kilogram  (Kg.). 

1000  kilograms  =  1  metric  ton  (T.). 

Table  of  Eqiiivalents 
English  to  Mktric  Metric  to  English 

1  oz.  Troy  =  31.1  g.  1  g.  =  0.0322  oz.  Troy. 

1  oz.  Av.    =  28.4  g.  1  g.  =  0.0353  oz.  Av. 

1  lb.  Av.     =  0.4.54  kg.  1  Kg.  =  2.2046  lb.  Av. 

1  ton  =  0.907  metric  ton.         1  metric  ton  =  1.10  tons. 

I.    Measurement  of  Angles 

60  seconds  (")  =  1  minute  (')•  t  =  3.14159265  •••. 

60  minutes         =  1  degree  (°).  =3.1416...  (nearly). 

90  degrees  =  1  right  angle.  =  31  (roughly). 

360  degrees  =  1  perigon  (circumference). 

1  radian  =  1^  =  ^  }^^l      =  57°  17'  45"  (nearly). 
TT         3.141b-" 

1°  =  •^•^^^^••'  radians  =  0.0175...  radians. 
180 


WEIGHTS   AND  MEASURES 


413 


J.   Measurement  of  Temperature 

The  unit  of  temperature  is  the  degree.  The  Centigrade  thermometer  is 
graduated  so  that  the  temperature  at  which  water  freezes  is  O'^,  and  the  tem- 
perature at  which  it  hoils  is  100°. 

The  Fahrenheit  thermometer  is  graduated  so  that  the  temperature  at  which 
water  freezes  is  ;}2°,  and  the  temperature  at  which  it  boils  is  212°. 

The  Reaumur  thermometer  is  graduated  so  that  the  temperature  at  which 
water  freezes  is  0°,  and  the  temperature  at  which  it  boils  is  80°. 

If  C  denotes  the  Centigrade  record  of  temperature  and  F  the  Fahren- 
heit, then, 


C=HF-=^-°)'  F=|C  +  32° 


(See  p.  143.) 


If  R  denotes  the  Reaumur  temperature, 

R  =  |C  =  HF-32°). 

A  comparison  of  these  three  scales  of  temperature  may  be  readily  seen 
from  the  following  table : 


Centigrade 

Faiikexiieit 

I'.EAIMIK 

Freezing 

Boiling 

From  boiling  to  freezing 

0 
100 
100 

32 

212 
180 

0 
80 

80 

m.    TABLES  OF  SIMPLE    FORMULAS  FROM  PHYSICS 

1.  Falling  Bodies. 

With  the  notation  s  =  space  passed  over  (in  ft.  or  cm.),  t  —  time  (in 
sec),  (•  =  velocity  (in  ft.  per  sec.  or  cm.  per  sec),  g  —  gravitational 
acceleration  =  32.16  ft.  per  sec.  per  sec.  =  981  cm.  per  sec.  per  sec. 

If  drojiped  from  rest : 

V  =  (jt,    s  =  \  gf\    v^  =  2gs. 

In  general,  v  —  Vq=  gt,    .«  —  Sq  =  5  gt'  +  v^t, 

if  Vq  =  initial  velocity,  Sq  =  initial  distance. 

2.  Lever,  or  Balance. 

dW  =  Dw.     See  Fig.  30,  p.  176. 

3.  Boyle's  Law :  Pressure  and  Volume. 

pv  =  constant.     See  p.  218. 
Other  formulas  given  as  needed  in  the  text. 


414 


TABLES 


IV.    TABLES  OF   GEOMETRIC  MENSURATION  FORMULAS 

(J  Triangle 

Area   =  ^  (base  x   altitude)  =  \  AB  •  CX 
=  ^  b  •  a,  where  h  =  base  =  A  B,    and 
a  =  altitude  =  CX. 
Sum  of  angles  =  ZA  +  ZB  +  ZC 

=  2  rt.  A  =  180°.  c 

Bight  Triangle 
ZB  =  rt.Z  =  90°, 
ZA  +  ZC  =  90°. 
(Hypotenuse)^  =  sum  of  squares  of  the 
perpendicular  sides  : 

AC^  —  AB'^  +  BC'^.  Right  Triangle 

If  ,1  C  =  h(=  hypotenuse),  AB  =h  (=  base),  CB  =  a(=  altitude), 
the  formula  becomes,  ^2  _  „2  _^  j2. 


A  X 

Triangle 


If 

and 
D 


Ratios  in  Right  Triangle 

-  =  s  (  =  sine  of  Z  .1 ),  -  =  c  (  =  cosine  of  Z  A), 
h  h 

-—  t  {=  tangent  of  ZA),  then: 
b 


-=  t 


Square 


1+4 


1  1 


1  + 


i- 


Square 
Area    =  (length  of  side)^  =  AB-  =  s^,  where 
s  =  AB  —  length  of  side. 


and 


Bectangle 
Area    =  base  x    altitude    =  AB  x  BC 
=  b  ■  a,    where  b  =  base  =    A  B, 
a  =  altitude  :^BC. 

D       X  C 


D 


C 


Rectangle 


U 


Y      B 
Parallelogram 


Parallelogram 

Area   =  base    x    altitude  =    AB  x  XY 
—  h  ■  a,  where  h  =  base  —  AB,  and 
where  a  =    altitude  —  XY. 


MENSURATION    FORMULAS 


415 


Circle 

Circuiiiference  =  2  tt  x  radius  =  2  7rr; 

Area  =  tt  x  (radius)^  =  irr% 

where  r  =  radius  of  circle, 

and,  see  p.  112,  tt  =  3.1416...  —  3|  (nearly). 


/ 

/ 

/ 

/ 

Pyramid 


Cube  Circle 

Volume  =  (length  of  .side)^  =  «'', 
where      s  =  length  of  side. 


Cube  Rpctanyrilar  Parallelopipcd 

Volume  = 
(area  of  base)  x  altitude. 

Pi/ramid  -  any  Base     rectangular  Parallelopiped 

Volume  = 
\  (area  of  base)  x  altitude. 

Regular  Pyramid 
(Base  a  regular  polygon ;  perpen- 
dicidar  from  vertex   meets   base   at 
center.) 
Lateral  area  =  \  (perimeter  of  base)  x 
(altitude  of  face). 

The  altitude  of  face  is  the  altitude  of     Regular 
any  one  of  the  triangular  faces,  often  called  Pyramid 

ttlaiit  height. 

Eight  Cylinder 

Lateral  area  =  (circumference  of  base)  x  altitude. 
Volume  =  (area  of  base)  x  altitude. 

Right  Cone 
Lateral  area  =  \  (circumference  of 
base)  X  slant  height  {i.e.  length  of  ele- 
ment) . 

Volume  =  \  (area  of  base)  x  altitude. 

Sphere 
Area       =  4  tt  x  (radius)"^  =  4  tt/-^. 
Sphere  Volume  =  5  tt  x  (radius) »  =  *  irr^.  Right  Cone 


INDEX 


(The  numbers  refer  to  pages,     v.  =  vide.) 


Abbreviations    (v.    Substitution    and 

Signs),  2. 
Addition,  1,  :M. 

axioms  of,  .'55,  56. 

of  expressions,  4!>.    ■ 

of  fractious,  125,  129. 

of  monomials,  44. 

of  negatives,  15,  ;>6,  39. 

of  radicals,  289,  388. 
Algebra,  1. 

Alternation  (in  proportion),  139. 
Answers  (v.  Solutions  and  Roots),  19, 
104,  109,  205. 

checking  of,  v.  Check. 

false,  108,  .300,  313. 

irrational,  200,  214. 

negative,  109. 
Approximations      (v.      Errors      and 

Graphs),  19,  30,  144. 
Arithmetic  (v.  special  headings),  1. 
Arithmetic  Sequences,  v.  Sequences. 
Associative  Law, 

of  addition,  35. 

of  multiplication,  35. 
Averages,  41. 
Axioms, 

of  addition  and  multiplication,  35. 

of  exponents,  285. 

"of  operations  on  equations,  56. 

Base  (of  logarithms),  341. 
Binomial,  8. 
Binomial  Theorem,  3*59. 
Boyle's  Law,  218. 
Braces,  v.  Parentheses. 
Brackets,  v.  Parentheses. 

Cancellation, 

in  equations,  153. 

in  fractions,  132. 
Centigrade,  v.  Thermometer. 


Chance,  3(!1. 
Changes,  Permissible, 

in  equations  (v.  Equations),  55,  106, 
137,  1.53. 

in  fractions,  119,  124. 

in  proportion,  137. 

in  rationalization,  .292. 
Characters,  v.  Signs. 
Characteristic  (of  logarithm.s),  343. 
Checks  (v.  Graphs),  5. 

complete,  5. 

in  addition,  etc.,  45,  78. 

in  English  problems,  5,  61. 

in  equations,  5,  57,  etc. 

in  radical  equations,  302. 

in  radicals,  287. 

in  substitution  methods,  313. 
Choice,  3(il. 
Circle,  250. 

Clearing  of  Fractions,  138,  149,  1.53. 
Clearing  of  Radicals,  292,  300,  304. 
Coefficients,  8. 

choice  of,  45. 

detached,  354. 

radical,  308. 

relation  to  roots,  224. 
Combinations,  3()4. 
Common  Factors,  v.  Factors. 
Common  Multiples,  v.  Multiples. 
Commutative  Laws,  35. 
Comparison,  solution  by,  172. 
Completing  a  Square,  206. 
Complex  Fractions,  134. 
Complex  Numbers,  391. 
Composition  (in  proportion),  137. 
Computation,  340,  .'547. 
Conjugate  Complex  Numbers,  392. 
Conjugate  Radicals,  294. 
Convergent  Series,  385. 
Cosine,  251,  414. 
Cross-multiplication  (in  proiX)rtion),  138. 


hedkiok's  el.  alg.  —  27       417 


418 


INDEX 


Cube  Root,  V.  Root. 
Curve,  V.  Graph. 

Degree, 

of  equation,  152,  255. 

of  radicals,  284. 

of  terms,  152,  255. 

reduction  of,  2V)0. 
Denominator  (v.  Fraction),  118. 
Detached  Coefficients,  354. 
Diagram,  v.  Graph. 
Discriminant,  214,  227,  229. 
Divergent  Series,  385. 
Division  (v.  Fractions),  1,  74. 

by  zero,  56,  75,  79,  106,  119,  150. 

exact,  84. 

in  proportion,  78. 

of  fractions,  74,  134. 

of  longer  expressions,  79,  84. 

of  monomials,  77. 

of  powers,  76. 

of  radicals,  286. 

Element  (of  a  sequence),  323. 

Elimination,  163. 

English  Statements  {v. Problems),  103. 

Equal,  2. 

Equal  Roots,  v.  Roots. 

Equality,  v.  Equations. 

Equations,  2. 

answers  to,  v.  Answers. 

cancellation  in,  153. 

cubic,  152. 

degree  of,  152,  255. 

equivalent,  168.  ' 

formation  of,  223. 

fractional  (v.  Proportion),  149. 

graph  of  (v.  Graph),  23. 

homogeneous,  397. 

indeterminate,  159,242. 

of  degree  1,  v.  Equations,  Linear. 

of    degree   2,  v.   Equations,    Quad- 
ratic. 

of  degree  ??,  152. 

simple,  v.  Equations,  Linear. 

solution  of,  v.  Ansioers,  and  Roots, 
and  Problems. 

Symmetrical,  317,  402. 
Equations,     Linear      (v.     Equations, 
Simultaneous  Linear,  and  Varia- 
tion, Linear),  25,  140,  l.">2,  239. 

graph  of,  V.  Graph. 


[Equations,  Linear], 

operations  on,  55,  106, 137,  153. 

solution  of,  57. 
Equations,    Quadratic,   109,    1.52,   203, 
253,  2.55,  396,  316. 

character  of  roots,  214. 

discriminant,  214,  227,  229. 

given  roots,  223. 

graph,  204.. 

literal,  229. 

relation  of  roots  to  coeflScients,  224. 

simultaneous,  267. 

solution   of,  109,  151,  205,  206,  209. 
212,  227,  261,  316,  3%. 
Equations,  Radical,  146,  318. 
Equations,   Simultaneous, 

Linear,  159,  160,   163,  170,  172,  173, 
319. 

Linear  and  Quadratic,  253,  319. 

Quadratic,  267,  276,  397. 
Errors  (v.  Apiproximations) ,  108. 
Euclidian  Method,  H.C.F.  and  L.C.M., 

373. 
Evolution,  181. 
Exponents  (v.  Logarithms),  8. 

fractional,  192,' 194,  285. 

negative,  29(). 

rules  for,  194,  285,  334. 

table  of,  335,  339. 

zero,  2i)(). 
Expression,  7. 

quadratic,  225. 

radical,  184,  192,  284. 

surd,  186,  284,  306. 

Factorial,  .364. 
Factors,  91. 

by  grouping,  100. 

common,  119. 

difference  of  powers,  99,  331,  358. 

difference  of  squares,  iH,  99. 

highest  common,  120,  373. 

monomial,  79. 

of  perfect  square,  93. 

polynomial,  91,  121,  128,  359. 

quadratic  expressions,  98,  107,  225. 

sum  of  cubes,  99. 

sum  of  powers.  99,  331,  358. 

theorem,  223,  .3.57. 

trinomial,  9(i,  98,  105,  107,  225. 

type  forms,  91,  .3.'>1,  358. 
Fahrenheit,  v.  TItermometer. 


INDEX 


419 


Falling  bodies,  221,  234. 
Fractions  (v.  Rdtio),  1,  118. 

additiiiu  of,  125,  liU. 

cleaiiiii;  of,  v.  Clearing. 

complex,  I'.'A. 

division  of,  74,  ISl. 

multiplication  of,  G7,  131. 

rational,  1«(>,  284. 

reduction  of,  118. 
Fractional  Equations,  v.  Eqxiations. 
Fractional  Exponents,  v.  ExponoiU. 
Function,  linear  (v.  Variation,  Linear), 
2:38. 

Geometric  Sequences,  328. 
Geometry,  formulas  of,  414. 
Graphs  (v.  Curves,  and  Straight  Line), 
19,  23. 

characteristic  of  logarithm,  343. 

cube  roots,  189s 

higher  roots,  VM. 

inequalities,  3()(i. 

linear  variation,  25,  140,  238. 

logarithms,  330,  338. 

of  prices,  20. 

plotting  of,  247. 

quadratics,  .204. 

raising  vertically,  25,  271. 

simultaneous  linear  equations,  15'.t.^ 

square  root,  182.'*t 

various,  248. 
Grouping,  v.  Parentheses. 

Highest  Common  Factor,  v.  Factor. 
Hindu  Method  (in  quadratics),  20'J. 
Homogeneous  Equations,  397. 

Imaginary  Numbers,  182,  210,  285,  390. 
Indeterminate  Equations,  v.  Equations. 
Index  of  root,  9. 
Inequalities,  ofitJ. 
Infinite  Series,  v.  Series. 
Integers,  1. 

Intersection,  point  of,  VV),  2."5. 
Inversion  (in  proportion),  139. 
Involution,  ISI. 

Irrational  Numbers  (v.  Surd),  186,  284, 
342,  38(i. 

Lever,  I>aw  of,  176. 

Limits,  ;'>S(). 

Line,  v.  Graph,  Curve,  Straight  Line. 


Linear  Equations,  v.  Equations. 
Logarithms  (v.  Exponents),  337. 

Briggs-,  342. 

common,  342. 

rules,  i{42. 

table,  339,  348. 
Lowest  Common  Multiple,  v.  Multiple. 
Lowest  Terms,  v.  Terms. 

Mantissa  (of  logarithms),  343. 
Marks,  v.  Sir/ns. 
Mathematical  Induction,  .')71. 
Measurement,  errors  in,  108. 
Measures,  4<)S. 
Mensuration  Formulas,  414. 
Metric  System,  408. 
Mirror,  Concave,  220. 
Monomials,  8. 

addition  of,  44. 

division  of,  77. 

multiplication  of,  71,  73. 

powers  of,  192. 

roots  of,  192. 

subtraction  of,  45. 
Multiples,  127. 

common,  127. 

lowest  coiumon,  128,  149,  375. 
Multiplication  (v.  Product),  1,  68. 

by  zero,  ()8,  10(),  109, 

of  fractions,  67,  131. 

of  longer  expressious,  78. 

of  negatives,  70. 

powers,  72. 

radicals,  28*;. 

Negative  Exponents,  295. 
Negative  Numbers,  14. 

addition  of,  15,  36,  39. 

as  answers,  109. 

averages  of,  41. 

multiplication  of,  70. 
Numbers,  1. 

complex,  391. 

conjugate  complex,  392. 

imaginary,  182,  210,  285,  390. 

irrational,  18(),  284,  342,  386. 

rational,  186,284,387. 

real,  284,  390. 
Numerator,  118. 

Operations,     in    equations,    etc.,     v. 

Changes,  ' 


420 


INDEX 


Parallel,  166. 
Parentheses,  7,  10,  47,  53, 
Permutations,  363. 
Physics,  formulas  of,  413. 
Polynomials,  8,  9,  88,  91, 121. 
Powers,  8,  181. 

divisiou  of,  76. 

fractional,  192,  194,  285. 

multiplication  of,  72. 

of  longer  expressions,  196. 

of  monomials,  192. 

of  radicals,  187. 

positive  integral,  8,  193. 

rules  for,  194,  285. 

simple,  8,  193. 
Prices, 

graph  of,  20. 

equation  of,  24. 
Problems  in  English, 

directions,  5,  59,  61,  103,  109. 

structure  of,  103. 
Product  (v.  Multiplication),  1,  8,  68. 

equal  to  zero,  107,  153. 

sum  and  difference,  i>4. 
Progressions,  v.  Sequences. 
Proportion  (v.  Equations,  Fractional 
and  Variation,  Linear),  2,  25, 137. 

between  variables,  139,  237. 

operations  in,  137. 

Quadratic  Equations,  v.  Equations. 
Quotient,  v.  Fraction  and  Division. 

Radicals  (v.   Surd   and  Irrationals), 
184,  192,  284. 

addition  of,  289, 

conjugate,  2M. 

degree  of,  284. 

division  of,  286. 

equations,  v.  Equations. 

expressions,  184,  192,  284. 

multiplic^ation  of,  286. 

operations  on,  188. 

powers  of,  187. 

rationalization  of,  292. 

reduction  of,  290. 

sign,  2,  192,  285. 

similar,  289. 

simple.st  form,  293. 
Ratio  (v.  Fraction),  VM,  237. 

common  (in  sequence),  .'528. 


[Ratio], 

in  right  triangle,  251,  414. 

of  proportionality,  237. 
Rational, 

fraction,  186,  284. 

number,  18(;,  284,  387. 
Rationalization  (of  radicals),  292. 

of  denominator,  292,  294. 
Real  Numbers,  284,  .390. 
Reaumur,  v.  Thermometer. 
Reciprocal,  l.'il,  296. 
Reductio  ad  absurdum,  166,  306. 
Remainder, 

in  division,  85. 
Remainder  Theorem,  360. 
Reversible  Process,  302. 
Roots,  8,  182. 

oube,  9,  189,  377. 

even,  182. 

higher,  190,  379. 

imaginary,  182,  210,285,  390. 

index  of,  9. 

longer  expressions,  19<j. 

monomials,  192. 

odd,  182. 

■square,  2,  104,  181,  184,  197,  305. 
Roots  of  an  Equation   (v.  Answers), 
205. 

equal,  209,  214. 

given,  223. 

imaginary,  210,  214. 

unequal,  214. 


Sequences  (v.  Series),  323. 

arithmetic,  282,  323. 

geometric,  328. 
Series  (v.  Seqxiences)  (convergent,  di- 
vergent, etc.),  .382. 
Signs,  1. 

rules  of,  71,75,  124. 

tables  of,  407. 
Similar,  v.  Terms  and  Radicals. 
Simple,  V.  Equations  and  Powers. 
Simultaneous,  v.  Equations. 
Sine,  2.51,  414. 

Solutions,  V.  Answers  and  Equations. 
Square.  4,  8,  30. 

conipletiiiga,  206. 

of  difference,  93. 

of  sum,  93. 
Square  Root,  v.  Root. 


INDEX 


421 


straight  Lines,  25,  140,  239. 

parallel,  Kjli. 

plotting,  142,  144. 
Structure  (of  problems),  103. 
Substitution,  45,  78,  91. 

method  of,  313. 

solution  by,  170,  25(). 
Subtraction  (v.  Addition),  1, 15,  39,  45, 

51,  12.->. 
Surds    (v.  Radicals  and  Irrationals), 
ISC,  214,  284. 

equality  of,   306. 

expressions,  186,  284. 

square  roots  of,  305. 
Symbols,  v.  Signs. 

Tables,  407  et  seq. 

Tangent  (to  a  curve),  259. 

Tangent  (ratio  in  right  triangle),  251, 

414. 
Temperature  (v.       Thermometer) 

(curves),  17. 
Terms  (of  an  expression),  7. 

like,  or  similar,  8,  45. 
Terms  (of  a  fraction),  118. 

lowest,  120,  122. 
Terms  (of  a  sequence),  323. 


Thermometer,  413. 

Centigrade,  13,  142. 

Fahrenheit,  13,  142. 

Keaunmr,  14(i. 
Transposition,  59,  153. 
Trial  Divisor,  185,  198,  378. 
Trinomial,  8. 

factors  of,  96,  98,  105,  107,  225. 

Variables,  25,  139,  237. 
Variation,  237. 

as  cube,  249. 

as  square,  248. 

inverse,  239. 

linear,  125,  140,  238. 

simple,  237. 

simultaneous,  238. 

various,  248  et  seq. 
Vinculum,  v.  Parentheses. 

Weights,  408. 

Zero,  15. 
division  by,  56,  75,  79,  106,  119,  150 
exponent,  296. 
multiplication  by,  68, 106,  109. 


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